Multiscale Modelling for Structures and Composites

MULTI-SCALE MODELLING FOR STRUCTURES AND COMPOSITES Multi-scale Modelling for Structures and Composites by G. PANASEN...

Author: G. Panasenko

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MULTI-SCALE MODELLING FOR STRUCTURES AND COMPOSITES

Multi-scale Modelling for Structures and Composites by

G. PANASENKO

A C.I.P. Catalogue record for this book is available from the Library of Congress.

ISBN 1-4020-2981-0 (HB) ISBN 1-4020-2982-9 (e-book)

Published by Springer, P.O. Box 17, 3300 AA Dordrecht, The Netherlands. Sold and distributed in North, Central and South America by Springer, 101 Philip Drive, Norwell, MA 02061, U.S.A. In all other countries, sold and distributed by Springer, P.O. Box 322, 3300 AH Dordrecht, The Netherlands.

Printed on acid-free paper

All Rights Reserved © 2005 Springer No part of this work may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, microfilming, recording or otherwise, without written permission from the Publisher, with the exception of any material supplied specifically for the purpose of being entered and executed on a computer system, for exclusive use by the purchaser of the work. Printed in the Netherlands.

Contents

Preface Notations Chapter 1. Introduction: Basic Notions and Methods 1.1. What is an inhomogeneous rod? 1.2. What are effective coefficients? 1.3. A scheme for calculating effective coefficients 1.4. Microscopic structure of a field 1.5. What is the homogenization method? 1.6. What is a finite rod structure? 1.7. What is a lattice structure? 1.8. Advantages and disadvantages of the asymptotic approach 1.9. Appendices 1.A1. Appendix 1: What is the Poincare-Friedrichs-Korn ´ inequalities? Chapter 2. Heterogeneous Rod 2.1. Homogenization (N.Bakhvalov’s ansatz and the boundary layer technique) 2.1.1. Bakhvalov’s ansatz 2.1.2. An example of formal asymptotic solution 2.1.3. The boundary conditions corrector 2.1.4. Introduction to the boundary layer technique 2.1.5. Homogenization in IRs 2.1.6. Boundary layer correctors to homogenization in IRs 2.2. Steady-state conductivity of a rod 2.2.1. Statement of the problem 2.2.2. Inner expansion 2.2.3. Boundary layer corrector 2.2.4. The justification of the asymptotic expansion 2.3. Steady state elasticity equation in a rod 2.3.1. Formulation of the problem 2.3.2. Inner expansion 2.3.3. Boundary layer corrector 2.3.4. The boundary layer corrector when the left end of the bar is free 2.3.5. The boundary layer corrector for the two bar contact problem 2.3.6. Homogenized problem of zero order 2.3.7. The justification of the asymptotic expansion 2.4. Non steady-state conductivity of a rod 2.4.1. Statement of the problem 2.4.2. Inner expansion

ix xii 1 1 3 5 9 9 10 13 16 17 17 21 22 22 23 26 27 29 32 36 36 38 43 48 56 57 59 65 69 73 79 82 98 98 98

VI 2.4.3 Boundary layer corrector 2.4.4. Justification 2.5. Non steady-state elasticity of a rod 2.5.1. Statement of the problem 2.5.2. Inner expansion 2.5.3. Boundary layer corrector 2.5.4. Justification 2.6. Contrasting coefficients (Multi-component homogenization) 2.7. EFMODUL: a code for cell problems 2.8. Bibliographical Remark Chapter 3. Heterogeneous Plate 3.1. Conductivity of a plate 3.1.1. Statement of the problem 3.1.2. Inner expansion 3.1.3. Boundary layer corrector 3.1.4. Algorithm for calculating the effective conductivity of a plate 3.1.5. Justification of the asymptotic expansion 3.2. Elasticity of a plate 3.2.1. Statement of the problem 3.2.2. Inner expansion 3.2.3. Boundary layer corrector 3.2.4. Proof of Theorem 3.2.1. 3.2.5. Algorithm for calculating the effective stiffness of a plate 3.3. Equivalent homogeneous plate problem 3.4. Time dependent elasticity problem for a plate 3.5. Bibliographical Remark Chapter 4. Finite Rod Structures 4.1. Definitions. L-convergence 4.1.1. Finite rod structure 4.1.2. L-convergence method for a finite rod structure 4.2. Shape optimization of a finite rod structure 4.2.1. Stored energy as the cost 4.2.2. Simplification of the set of finite rod structures. Initial configuration 4.2.3. An iterative algorithm for the optimal design problem 4.2.4. Some results of numerical experiment 4.3. Conductivity: an asymptotic expansion 4.3.1. Construction of asymptotic expansion 4.3.2. The leading term of the asymptotic expansion 4.4. Elasticity: an asymptotic expansion 4.4.1. Construction of asymptotic expansion 4.4.2. The leading term of asymptotic expansion 4.5. Flows in tube structures

100 100 103 103 104 106 106 110 120 126 129 130 130 131 134 136 138 146 146 146 149 151 154 156 158 160 161 161 162 165 173 174 176 177 180 183 183 192 194 194 212 215

VII 4.5.1. Definitions. One bundle structure 4.5.2. Tube structure with m bundles of tubes 4.6. Bibliographical Remark 4.7. Appendices 4.A1. Appendix 1: estimates for traces in the pre-nodal domain 4.A2. Appendix 2: the Poincaré and the Friedrichs inequalities for a finite rod structure 4.A3. Appendix 3: the Korn inequality for the finite rod structures Chapter 5. Lattice Structures 5.1. Definition of lattice structure 5.2. L-convergence homogenization of lattices 5.2.1. L-convergence for the simplest lattice 5.2.2. Some auxiliary inequalities 5.2.3. FL-convergence. Relation to the L-convergence 5.2.4. Proof of Theorem 5.2.1 5.3. Non-stationary problems 5.3.1. Rectangular lattice 5.3.2. Lattices: general case 5.3.3. Proof of Theorem 5.3.2 5.4. L- and FL-Convergence in elasticity 5.5. Conductivity of a net 5.6. Elasticity of a net 5.7. Conductivity of a lattice: an expansion 5.8. High order homogenization of elastic lattices 5.9. Random coefficients on a lattice 5.9.1. The Simplest Lattice. The main result 5.9.2. Proof of theorem 5.9.1 5.10. Bibliographical Remark 5.11. Appendices 5.A1. Appendix 1: the Poincaré and the Friedrichs inequalities for lattices 5.A2. Appendix 2: the Korn inequality for lattices Chapter 6. The Multi-Scale Domain Decomposition 6.1. Differential version 6.1.1. General description of the differential version 6.1.2. Model example 6.1.3. Poisson equation in a rod structure 6.2. Variational version 6.2.1. General description of the variational version 6.2.2. Model example 6.2.3. Elasticity equations

215 227 230 230 230

233 242 247 247 249 249 251 254 255 258 258 259 262 269 270 273 278 291 307 307 309 314 318

318 321 337 341 341 343 349 354 354 357 358

VIII 6.3. Decomposition of a flow in a tube structure 6.4. The partial homogenization 6.5. Bibliographical Remark

364 373 382

Bibliography Subject index

385 397

PREFACE. Rod structures are widely used in modern engineering. These are bars, beams, frames and trusses of structures, gridwork, network, framework and other constructions. A variety of theories based on the Kirchhoff-Love, KirchhoffClebsch and other hypotheses are applied for their analysis. Structural mechanics software based on material strength theory methods also exists. At the same time the questions concerning the limits of applicability of these hypotheses and theories and the possibilities of their refinement are very important. In this connection we develop the multi-scale asymptotic analysis of equations of mathematical physics, and in particular the elasticity equations of set in the rod structures (without these hypotheses and simplifying assumptions being imposed) . Problems with one small parameter (the ratio of bar diameter to its length) as well as problems with two and more small parameters (periodic framework systems, where the second parameter represents the ratio of a period to the characteristic space dimension of the problem, weakly compressible bars, etc.) are studied. The homogenization technique for partial differential equations described in the book by N.Bakhvalov and G.Panasenko [16] and the boundary layer techniques are used as a main tool in these investigations. The physical processes are simulated by partial differential equations set in ”thin” domains containing a small parameter. The asymptotic analysis is applied for investigation of these partial differential equations. The multi-scale models are developed according to two main schemes: - the up-scaling procedure of asymptotic derivation of macroscopic models from the microscopic ones (ie., the homogenization approach) and - the hybrid multi-scale models, combining two scales inside one model, making a microscopic zoom inside the macroscopic model (ie., the asymptotic partial domain decomposition, partial homogenization). The present monograph consists of six chapters. The first chapter is introductory. It presents the main notions and methods of the book, advantages and disadvantages of these methods. In the second chapter we consider the three-dimensional conductivity problem as well as theory for elasticity problems and other equations of mathematical physics in a thin cylindrical domain (bar) with non-homogeneous structure. Full asymptotic expansions of solutions are constructed for the small parameter equal to the ratio of the bar diameter to the length; boundary layers are investigated and one-dimensional equations for bars are derived. Then we consider the problem of junction of two heterogeneous rods. The connection conditions for rods result from the analysis of boundary layers arising in the neighborhood of the bounds of the rods. Time dependent models are considered. The third chapter is devoted to the similar analysis of a heterogeneous layer (plate). Full asymptotic expansions of solutions are constructed for the small parameter equal to the ratio of the plate thickness to the length of the plate; boundary layers are studied and two-dimensional equations for plates are deIX

X rived. Similar questions for systems with finite number of bars are studied in the Chapter 4. The asymptotic analysis of a structure with finite number of bars is developed first for nonlinear equations in some weak norms (L-convergence method), and then the detailed asymptotic analysis is done, the asymptotic expansions of solutions are constructed. The Korn inequality or, more specifically, the investigation of how the constant depends on the small parameters in this inequality is essential for the justification. Finally the results obtained by means of L-convergence method are applied to shape design of finite rod structures. The code OPTIFOR implements this algorithm. We discuss some model examples. The similar analysis is developed for some problems of flow in a system of tubes described by Stokes and Navier-Stokes equations. In Chapter 5, periodic framework structures (lattice-like domains) are considered and their homogenization is carried out, i.e., systems with great number of bars are investigated. The question of the existence of homogenized models and the convergence of the exact solutions to them is studied. In some cases the first approximation of the asymptotic theory may be also obtained by means of classical structural mechanics. At the same time, the asymptotic theory gives the opportunity to obtain corrections to the structural mechanics models. It is possible to obtain analytical formulas for these corrections in some examples. Chapter 6 is devoted to a new multi-scale method of solution of different problems with small parameter. It is the method of asymptotic domain decomposition. The direct numerical solution of partial derivative equations in finite rod structures is very expensive because the complicated geometry demands a large number of nodes in the grid. The complete asymptotic expansions are often cumbersome. So we propose a hybrid numerical-asymptotic method which uses a combined 3D-1D models: it is three-dimensional in the boundary layer domain and it is one-dimensional outside of the boundary layer domain. We cut the rods at some distance from the ends of the rods, we keep the dimension three in the neighborhood of the ends and we reduce dimension on the truncated (main) part of rods. So the principal idea of the method is to extract the subdomain of singular behavior of the solution and to reduce dimension of the problem in the subdomain of regular behavior of the solution. Of course the most important question is: what are the interface conditions between 3D and 1D parts? We formulate two approaches of construction of such hybrid models and justify the closeness of the partially decomposed model and initial model. We analyze such hybrid models for conductivity and elasticity equations stated in rod structures as well as Stokes and Navier-Stokes equations stated in a system of thin tubes. We consider below two versions of the method of asymptotic partial decomposition of domain. The first version is ”differential”, i.e. we work with the differential formulation of the initial problem , we obtain the 1D differential equation in the reduced part of the rod structure and we add the differential interface conditions on the boundary between 3D and 1D parts . Of course we can pass to a variational formulation of the partially decomposed problem but it is generated by a differential one.

XI The second version is a direct variational approach, when the 3D integral identity for the original problem is restated for a special subspace of functions having a form of the ansatz of the asymptotic solution in the regular thin part of the rod structure. We give some examples of application of this method. As a rule the results are presented in the form of theorems. The obtained asymptotic approximations are justified: estimates of their closeness to exact solutions are proved. The book is destined to graduate and postgraduate students, specialists in applied mathematics, specialists in mechanics, engineers, university professors. It is partially based on courses of lectures delivered by the author at the Department of Mechanics and Mathematics at Moscow State University Lomonossov and at the Mathematical Department of the University Jean Monnet (Saint Etienne, France). Knowledge acquired after the first three years of advanced mathematical training at a college or university is sufficient to read the most of the book. Presentation of the material is based on the principle ”from simple to complicated,” every chapter beginning with an elementary example to illustrate the main idea of the method to be described.

XII Notations

ε, µ, ω −1 -small parameters G - a domain in IRs , i.e., an open connected set (usually bounded) in the s−dimensional space where s is equal to 2 or 3 ∂G -boundary of the domain G ¯ -closure of the domain G, i.e., G ¯ = G ∪ ∂G G Bµ -finite rod structure i.e., a connected union of a finite number of thin cylinders (the diameter of the base is of order of µ and the height is of order of 1); e is a segment inside of a cylinder constituting Bµ , concluded between the bases Bε,µ -lattice structure i.e., a connected ε−periodic union of an infinite number of thin cylinders (the diameter of the base is of order of product εµ and the height is of order of ε) x = (x1 , ..., xs ) -point in IRs , slow (macroscopic) variable ξ = (ξ1 , ..., ξs ) -fast (microscopic) variable, ξ = x/ε or ξ = x/µ z = (z1 , ..., zs ) -fast (microscopic) variable in a boundary layer, z = x/µ or z = x/(εµ) L2 (G) (the same that L2 (G)) space of functions with bounded norm uL2 (G) = u( x)dx G ¯ -space of functions continuous in G, ¯ uC(G) C(G) ¯ = supx∈G ¯ |u(x)| n ¯ ¯ C (G) -space of functions n times differentiable in G

H 1 (G), H K (G), H01 (G) -are the classical Sobolev spaces u|∂G -trace on ∂G [u]|Σ -means the difference of the limit values of function u on the two sides of surface Σ f (x1 , ..., xs , ξ1 , ..., ξs ) is a mean over the period, i.e.,

0

1

...

1

f (x1 , ..., xs , ξ1 , ..., ξs )dξ1 ...dξs

0

(a, b) -inner product of the vectors a and b div, divx , divξ -divergence with respect to the variables x or ξ grad, gradx , gradξ -gradient with respect to the variables x or ξ ∇, ∇x , ∇ξ -the same rot, rotx , rotξ -curl with respect to the variables x or ξ ∆, ∆x , ∆ξ -Laplacian with respect to the variables x or ξ ˜ I(ξ) - matrix of rigid displacements (translations and rotations) i = (i1 , ..., il ) is a multi-index, |i| = l is its length, ij ∈ {1, ..., s} l Di = ∂xi ∂...∂xi (partial derivative) 1 l ∞ f (x, ε) ∼ j=0 εj gj (x, ε) is an asymptotic expansion (a.e.) of f (x, ε), i.e., for any real N there exists M such that, for all m ≥ M the relation holds:

XIII

f (x, ε) −

m

εj gj (x, ε) = O(εN )

j=0

u, uµ , uε , uε,µ -exact solution of the problem (which depends on some parameters) (∞) (∞) (∞) u(∞) , uµ , uε , uε,µ -formal asymptotic solution (f.a.s.) which is normally proved to be an asymptotic expansion of the exact solution of the problem (K) (K) (K) u(K) , uµ , uε , uε,µ - the asymptotic approximation of order K v or ω the formal asymptotic solution of the homogenized problem ˆ. - effective coefficient symbol δij - the Kronecker symbol

Chapter 1

Introduction: Basic Notions and Methods 1.1

What is an inhomogeneous rod?

We define an inhomogeneous rod as a long cylinder constituted of a unidirectional chain of recurrent elements (cells). Each cell consists of some subdomains occupied by different materials: the inclusions of one compound and the matrix of another compound which fills the space between the inclusions . The dimensions of the cell are of the same order as the diameters of the subdomains. Let L be the length of the rod and µ the length of the recurrent cell. We assume that the diameter of the cross section of the cylinder is of the same order as the length of the recurrent cell and that µ 0),

f0 = f.

l=3 |i|=l

Thus the construction of the formal asymptotic solution by means of the homogenization method (N. Bakhvalov’s version) consists in substituting series (2.1.2) into equation (2.1.1), collecting similar terms, and putting the rapidly

32

CHAPTER 2. HETEROGENEOUS ROD

oscillating coefficients Hi (ξ) of the derivations Di v equal to constants hi . These constants are determined from the problems that give the solvability conditions for Ni . In the monograph by N. Bakhvalov and G. Panasenko [16], the justification of the formalism presented above is carried out and estimates for the difference between the partial sums of the series (2.1.2) and the exact solution are obtained for the case in which (2.1.1) is an elliptic equation or the system of equations of elasticity theory.

2.1.6

Boundary layer correctors to homogenization in IRs

The formal asymptotic solution for the simplest boundary value problem was first constructed by the author [132],[133] (see also [16],[100]). Consider equation (2.1.1) in the layer x1 ∈ (0, b), where b/ε is a positive integer and b is a number of the order of 1. Let the boundary conditions

u = g0 (x′ ), u = g1 (x′ ) (2.1.47) x1 =0

x1 =b

be set. Here and below for any s-dimensional vector a we set a′ = (a2 , . . . , as )∗ , i.e. a′ is a column vector comprising all components of a except for the first component. Suppose that the right-hand sides f , g0 , and g1 are T -periodic with respect to their arguments (T is a number of the order of 1 such that T /ε is a positive integer). The solution is also sought in the class of T -periodic functions of x2 , . . . , xs . The formal asymptotic solution to problem (2.1.1), (2.1.47) is sought in the form of Bakhvalov’s series uB with the corrector u0p + u1p of boundary-layer type, u(∞) = uB + u0p + u1p ,

(2.1.48)

where uB is defined by equation (2.1.2) and u0p =

∞

εl

l=0

u1p =

∞ l=0

εl

|i|=l

|i|=l

Ni0 (ξ)Di v(x) ξj =xj /ε ,

Ni1 (ξ)Di v(x) |ξ1 =(x1 −b)/ε,ξj =xj /ε,

j=2,...,s.

(2.1.49)

(2.1.50)

Substituting (2.1.48) into (2.1.1) and (2.1.47) and taking into account the fact that uB is as constructed above, we obtain ∞

εl−2

l=0

∞ l=0

εl

|i|=l

|i|=l

¯ i x Di v = f (x), H ε

′

¯i 0, x Di v = g0 (x′ ), N x1 =0 ε

(2.1.51)

(2.1.52)

2.1.

33

HOMOGENIZATION ∞ l=0

where

εl

|i|=l

′

¯i b , x D i v = g1 (x′ ), N x1 =b ε ε

(2.1.53)

′ ¯ i x = Hi x + Hi0 x + Hi1 x1 − b , x , H ε ε ε ε ε x − b x′ x x x 1 ¯i , + Ni1 = Ni + Ni0 N ε ε ε ε ε

Ni (ξ) and Hi (ξ) are the 1-periodic functions constructed above, and Nir (ξ) and Hir (ξ) are 1-periodic functions of ξ2 , . . . , ξs exponentially stabilizing to zero as ξ1 → ±∞ (+ for r = 0, − for r = 1). Here Hir (ξ) = Lξξ Nir1 ...il +

+Ai1 j (ξ)

∂ Nir2 ...il + Aki1 (ξ)N ∂ξk

∂N Nir2 ...il + Ai1 i2 (ξ)N Nir3 ...il , ∂ξξj

(2.1.54)

where i = (i1 , . . . , il ) with negative length |i| is assumed to be zero. Denoting Tir =

∂N Nir2 ...il ∂ + Ai1 i2 (ξ)N Nir3 ...il , Aki1 (ξ)N Nir2 ...il + Ai1 j (ξ) ∂ξξj ∂ξk

one can reduce (24) to the form Hir (ξ) = Lξξ Nir + Tir (ξ). Let us arrange Nir in a sequence as was done above for Ni . Namely, let r Ni ≻ Njr whenever |i| > |j|. Note that Tir is defined by the functions Njr with priority |j| < |i|. This condition may be used for the recurrent definition of Nir ¯i (0, ξ ′ ), in terms of Njr with lower priority if certain requirements on Hir (ξ), N ′ ¯ and Ni (b/ε, ξ ) are imposed. ¯i (0, ξ ′ ) = const, and N ¯i (b/ε, ξ ′ ) = const. Note Let Hir (ξ) = 0, r = 0, 1, N that, according to the construction of the previous subsection, Hi (ξ) = hi = ¯i (0, ξ ′ ) by h0 and N ¯i (b/ε, ξ ′ ) by h1 . If N r and hr const. Denote the constants N i i i i satisfying these conditions are found, then problem (2.1.51)–(2.1.53) is reduced to the following problem with constant coefficients for v: ∞

εl−2

l=2

hi Di v = f (x),

|i|=l ∞

εl

l=0

∞ l=0

|i|=l

εl

|i|=l

x1 ∈ (0, b),

h0i Di v x1 =0 = g0 (x′ ),

h1i Di v x1 =b = g1 (x′ ).

(2.1.55)

(2.1.56)

(2.1.57)

34


This problem can be solved by the standard technique of expanding v into a regular series (2.1.41). This again gives (2.1.46) for vq with the boundary conditions

h0∅ vq x =0 = g0q (x′ ), h1∅ vq x =b = g1q (x′ ), (2.1.58) 1 1

q where gr0 = gr , grq = − l=1 |i|=l hri Di vq−l x1 =0,b (q > 0). Hence, in order to construct the formal asymptotic solution to the original problem, we only need to construct functions Nir and constants hri satisfying the equations Hir (ξ) = 0, r = 1, 2, ¯i (0, ξ ′ ) = h0i + O(e−c/ε ), N ¯i b , ξ ′ = h1i + O(e−c/ε ), N ε where c is a positive constant. The functions Nir and the constants hri are constructed by induction. Namely, we set N∅r = 0 and hr∅ = 1 or I in the vectorial case. Let all Nir of priority |i| ≤ k be found. Let us define Tir (ξ) for all i with |i| = k + 1. Choose a constant h0i (|i| = k + 1) such that the 1-periodic in ξ2 , . . . , ξs solution to the problem

Lξξ Ni0 + Ti0 (ξ) = 0, Ni0 (0, ξ ′ )

ξ1 > 0, ′

= −N Ni (0, ξ ) +

h0i ,

(2.1.59) (2.1.60)

exponentially tending to zero as ξ1 → +∞, exists, and choose a constant h1i such that the similar solution to the problem Lξξ Ni1 + Ti1 (ξ) = 0,

ξ1 < 0,

Ni (0, ξ ′ ) + h1i , Ni1 (0, ξ ′ ) = −N

(2.1.61) (2.1.62)

exponentially tending to zero as ξ1 → −∞, exists (the possibility of such a choice is assumed). The possibility of such a choice in the case of the elliptic equation is proved [89], [90], and for the system of equations of elasticity theory this is proved in [129], [130]. Note that Ni (0, ξ ′ ) = Ni (b/ε, ξ ′ ), since b/ε is an integer and Ni is 1-periodic in ξ1 . In this connection, condition (2.1.62) is equivalent to the condition b Ni1 (0, ξ ′ ) = −N Ni , ξ ′ + h1i . ε

To determine the constants h0i , we first solve the problem i0 + Ti0 (ξ) = 0, Lξξ N

ξ1 > 0,

i0 (0, ξ ′ ) = −N N Ni (0, ξ ′ )

in the class of bounded functions 1-periodic in ξ ′ . Let the solution to this problem stabilize to a constant Cr0 (this is true for elliptic equations). Next, let 0 , and then h0i = − limξ1 →+∞ N i i0 − Ni0 = N

i0 , lim N

ξ1 →+∞

|i| = k + 1

2.1.

35

HOMOGENIZATION

is the solution to problem (2.1.59), (2.1.60). Problem (2.1.61), (2.1.62) can be solved in a similar manner. The induction step is complete. Remark 2.1.2. By construction, we obtain u(∞) asymptotically satisfying equation (2.1.1). At the same time, the boundary conditions are satisfied asymptotically not for u(∞) , but for uB + u0p at the left end and for uB + u1p at the right end. But in view of the exponential decay of Nir (ξ) for any N we have

(u(∞) − uB − u0p ) x =0 = u1p x =0 = O(εN ) 1

and similarly

1

(u(∞) − uB − u1p ) x =b = u0p x =b = O(εN ). 1 1

Thus, u(∞) is a formal asymptotic solution to problem (2.1.1), (2.1.47). The justification of the presented formalism is carried out in [16]. Remark 2.1.3. (On the justification procedure). In this section 2.1 we have constructed several asymptotic solutions to some problems depending on small parameter ε. It means that having a problem Lε uε = fε

(2.1.63)

stated for an unknown function uε with an ε−dependent linear operator Lε and right hand side fε (which stands for the data of this problem), we have (K) constructed (instead of this unknown uε ) an asymptotic approximation uε (K) such that it satisfies (2.1.63) with some remainder rε of order O(εK ), i.e. (K) K rε Hd = O(ε ). Here Lε : Hs → Hd , Hs , Hd are some normed spaces. It means that Lε uε = fε + rε(K) . (2.1.64) Assume that for any fε ∈ Hd problem (2.1.63) has a unique solution and that an a priori estimate holds: uε Hs ≤ Cffε Hd

(2.1.65)

with a constant C independent of fε and ε. Then subtracting (2.1.63) from (2.1.64) and applying estimate (2.1.65) to (K) (K) (K) the difference uε − uε (that is a solution to equation Lε (uε − uε ) = rε ) we get u(K) − uε Hs ≤ Crε(K) Hd , ε i.e., u(K) − uε Hs = O(εK ). ε

(2.1.66)

This is the main idea of justifications of asymptotic procedures in this book. Sometimes the constant C in (2.1.65) depends on ε in such a way that there exists α > 0 such that C ≤ C0 varepsilon−α where C0 does not depend on ε. Then we get

36


− uε Hs = O(εK−α ). u(K) ε

(2.1.66′ )

instead of (2.1.66). If estimate (2.1.66) holds true for any K and α is a positive integer number and if K u(K+α) − u(K) ε ε Hs = O(ε ),

(2.1.67)

then estimate (2.1.66’) can be improved. Indeed from (2.1.66) for K + α we get u(K+α) − uε Hs = O(εK ). ε

(2.1.68)

Comparing (2.1.67) to (2.1.68)and applying the triangular inequality we get estimate (2.1.66). If Lε is not linear then we must try to prove that for any f1ε , f2ε ∈ Hd problems Lε uiε = fiε , i = 1, 2, have a unique solution and that there exists a constant C independent of ε and fiε , such that, u1ε − u2ε Hs ≤ Cf1ε − f2ε Hd . The spaces Hs and Hd may also depend on ε. These ideas will be applied in particular in the next section, as well as further in the book.

2.2

Steady-state conductivity of a rod

2.2.1

Statement of the problem

Definition 2.2.1. Let β be a bounded domain in IR2 with piecewise smooth boundary, and let Uµ be the cylinder {x1 ∈ IR, x′ /µ ∈ β}, satisfying the strong cone condition [55]. The cylinder Cµ = Uµ ∩ {x1 ∈ (0, b)} is called a threedimensional (s = 3) rod. Here µ > 0 is a small parameter. We recall that x′ = (x2 , x3 ). Definition 2.2.1.′ In the two-dimensional case (s = 2) Uµ = IR×(−µ/2, µ/2), and the rectangle Cµ = Uµ ∩ {x1 ∈ (0, b)} = (0, b) × (−µ/2, µ/2) is called a twodimensional rod. Next, for s = 2, we assume that β = (−1/2, 1/2).

2.2.

37

STEADY-STATE CONDUCTIVITY OF A ROD

Figure 2.2.1. A rod (three-dimensional and two-dimensional ) We consider the stationary conductivity equation Pu ≡

s ∂ x ∂u = Fµ (x), Aij µ ∂xj ∂xi

i,j=1

x ∈ Cµ

(2.2.1)

with the boundary conditions

s x ∂u ∂u ni = 0, Aij ≡ µ ∂xj ∂ν i,j=1

u = 0,

x1 = 0

x ∈ ∂U Uµ ,

or x1 = b.

(2.2.2) (2.2.3)

Here (n1 , . . . , ns ) is a vector normal to the boundary Uµ and Aij (ξ) are functions satisfying the following conditions: 1)for any ξ ∈ IRs , Aij (ξ) = Aji (ξ), and 2) there exists κ > 0 such that, for any ηi ∈ IR, i = 1, ..., s, the following inequality holds: s

i,j,k,l=1

Aij (ξ)ηi ηj ≥ κ

s

(ηi )2 ,

κ > 0.

i=1

It is assumed that the functions Aij (ξ) are 1-periodic in ξ1 , infinitely differentiable everywhere outside a set Σ (consisting of smooth nonintersecting surfaces Σl ) up to this set Σ; we assume that there is only a finite number of surfaces Σl that intersect with the cylinder [0, b] × β; Σ ∩ ([0, 1] × ∂β) = ∅, and on Σ the coefficients have jump discontinuities. If Aij (ξ) depend on ξ1 , we require that b/µ be an integer. The right-hand side Fµ has the following structure: x Ψ(x1 ); Fµ = F µ

here F (ξ) is a 1-periodic matrix function of ξ1 , which has the same smoothness as the coefficients (see above) and that ψ ∈ C ∞ ([0, b]). On the interfaces of discontinuity of the coefficients, the natural conjugation conditions are assumed:

38


[u] = 0;

s

i,j=1

Aij ∂u/∂xj ni = 0,

(2.2.4)

where (n1 , . . . , ns ) is the normal vector to the interface of discontinuity, [·] is the jump of a function on this interface. For s = 3, problem (2.2.1)–(2.2.3) models the temperature field in a rod of length b with cross-section βµ = {x′ ∈ R2 , x′ /µ ∈ β}; the ends of the rod have the temperature zero and the lateral surface is insulated. The rod has an inhomogeneous structure: the elements of the conductivity tensor Aij are µ-periodic functions of the longitudinal coordinate x1 , they also depend on the transverse coordinates x′ . The right-hand side (sources) is the product of the fast-oscillating function F (x/µ) by the slowly varying factor ψ(x1 ). For s = 2, the problem models a two-dimensional temperature field in a plate of length b and thickness µ. We construct formal asymptotic solutions (f.a.s.) to problem (2.2.1)–(2.2.3) in two stages, as in section 2.1. In the first stage we construct f.a.s. to equation (2.2.1) with condition (2.2.2), using an analog of N. Bakhvalov’s series; on the second stage we construct the boundary layer corrector. For the reader’s convenience, we first establish a formalism for each stage and then justify the asymptotic behavior, i.e., prove different assertions stated in constructing f.a.s. Furthermore, we prove that f.a.s. is an asymptotic expansion of an exact solution to the problem and establish an error estimate for the partial sum of the series of f.a.s.

2.2.2

Inner expansion

Let us represent the right-hand side F as the sum F = F¯ + F,

(2.2.5)

where F¯ = F , F = 0. From now on in this section F (ξ, x) denotes the average F (ξ, x)dξ . mes β (0,1)×β

We seek f.a.s. in the form of an analog of Bakhvalov’s ansatz, u(∞) =

∞

µl Nl

l=0

∞ x dl ψ(x ) x dl ω(x ) 1 1 , + µl+2 Ml l l µ µ dx dx1 1 l=0

(2.2.6)

where Nl , Ml are 1-periodic in ξ1 functions, ω(x1 ) is a C ∞ - function. Substituting series (2.2.6) into (2.2.1), (2.2.2), and (2.2.4) and collecting terms with like powers of µ yields P u(∞) − F ψ =

∞ l=0

µl−2 HlN (ξ)

∞

dl ω l M dl ψ + µ Hl (ξ) l l dx1 dx1 l=0

− F ψ − Fψ, (2.2.7)

2.2.

39


where HlN (ξ) = Lξξ Nl + TlN (ξ), s ∂ ∂ , Lξξ = Ajm ∂ξm ∂ξξj j,m=1

TlN (ξ) =

s s ∂N Nl−1 ∂ (Aj1 Nl−1 ) + A1j + A11 Nl−2 ; ∂ξξj ∂ξξj j=1 j=1 ∞

∞

l=0

l=0

(2.2.8)

dl ψ dl ω l+1 M ∂u(∞) + µ Gl (ξ) l , = µl−1 GN l (ξ) l ∂ν dx1 dx1

GN l =

s s

m=1

j=1

Amj

∂N Nl + Am1 Nl−1 nm ; ∂ξξj

(2.2.9)

(2.2.10)

HlM , TlM , GM l are obtained by replacing N with M . Here, as in section 2.1, Nl = 0 whenever l < 0; N0 = 1. We require that, as in the procedure from section 2.1, HlN (ξ) = hN l ,

HlM (ξ) = hM l ,

l ≥ 0,

H0M (ξ) = F(ξ),

l > 0,

M where hN are constants. Moreover, we require that GN l and hl l (ξ) = 0, M Gl (ξ) = 0 on the surface (0, 1) × ∂β and that [N ] = 0, [GN ] = 0; [M ] = 0, l [GM ] = 0 on the discontinuity interfaces Σ of the coefficients A (ξ). We obtain ij l the following recurrent chain of problems for Nl , Ml :

Lξξ Nl = −T TlN (ξ) + hN l , ∂/∂ννξ Nl = −

[N Nl ] Σ = 0,

s

m=2

Am1 Nl−1 nm ,

ξ ∈ (0, 1) × β,

(2.2.11)

ξ ∈ (0, 1) × ∂β,

(2.2.12)

s

∂N Nl /∂ννξ Σ = − Am1 Nl−1 nm Σ .

(2.2.13)

m=1

Figure 2.2.2. A periodic cell Here, as in section 2.1, hN l are chosen from the solvability conditions for (2.2.11), (2.2.12), (2.2.13) (see Lemma 2.2.1): (−T TlN + ΦhN l ) = −

s

m=2

Am1 Nl−1 nm (0,1)×∂β −

s

m=1

[Am1 Nl−1 nm ]Σ ,

40


where for the (s − 1)-dimensional hyper-surface Γ we have dξ , ·Γ = Γ mes β

i.e., hN l =

s

A1j

j=1

∂N Nl−1 + A11 Nl−2 . ∂ξξj

(2.2.14)

Ml are the solutions of the same problems with Nl replaced by Ml . However, M0 is the solution to the problem Lξξ M0 = F,

∂M M0

= 0,

∂ννξ ξ′ ∈∂β

[M M0 ] Σ = 0,

∂M M0

= 0. (2.2.15) ∂ννξ Σ

Thus, the algorithm for constructing the matrices Nl and Ml is inductive. Suppose that Nl = 0, Ml = 0 for l < 0, N0 = 1, and M0 is the solution to problem (2.2.15). If l > 0, then Nl and Ml are the solutions to problems (2.2.11)–(2.2.14). The right-hand sides of these problems contain Nm and Mm with indices m < l, which permits one to define them successively. We have hM 0 = 0,

hN 0 = 0,

hN 1 = 0,

hN 2 =

s j=1

A1j

∂N N1 + A11 . ∂ξξj

(2.2.16)

Here a 1-periodic solution Nl to problem (2.2.11)-(2.2.13) is understood in 1 the following sense. Denote Q = (0, 1) × β. Let Hper ξ1 (Q) be the completion 1 in the norm of H (Q) of the space of 1-periodic in ξ1 infinitely differentiable functions f (ξ) defined at IR × β. 1 A solutionU ∈ Hper ξ1 (Q) to the problem Lξξ U = F0 (ξ) +

s ∂F Fj j=1

s ∂U = Fj nj , ∂ννξ j=2

∂ξξj

,

ξ ∈ IR × β,

ξ ∈ IR × ∂β,

is understood as a function U satisfying the integral identity

s s ∂Ψ ∂U ∂Ψ = F F0 Ψ − Fj − Aij ∂ξξj ∂ξξj ∂ξi j=1 i,j=1 1 for any function Ψ ∈ Hper ξ1 (Q). Here the right-hand sides F0 , F1 , . . . , Fs belong to L2 ((−A, A) × β) for each A and are 1-periodic in ξ1 . Lemma 2.2.1. This problem has a solution if and only if

F F0 = 0.

2.2.

41


Proof. The proof is based on the Riesz representation theorem for a bounded linear functional on a Hilbert space. Namely, the right-hand side of the integral identity represents the linear functional G(Ψ) that is bounded in the norm of H 1 (Q) by the inequality |G(Ψ)| ≤

s j=0

F Fj L2 (Q) ΨH 1 (Q) .

On the other hand, for functions Ψ ∈ H 1 (Q) that satisfy Ψ = 0, the 2 s ∂Ψ ≥ c1 Ψ2H 1 (Q) , c1 > 0, holds (cf. [85]) Poincar´ é inequality i=1 ∂ξ i

Since

s

∂Ψ ∂Ψ Aij ∂ξ ξj ∂ξi i,j=1

the norm

≥ κ

2 s ∂Ψ i=1

∂ξi

,

s ∂Ψ ∂Ψ Aij Ψ1 = ∂ξξj ∂ξi i,j=1

1 is equivalent to the H 1 (Q)-norm on the subspace of functions of Hper vanishing average and this norm corresponds to the inner product

[Ψ, Θ]1 =

ξ1 (Q)

with

∂Ψ ∂Θ . Aij ∂ξξj ∂ξi i,j=1

s

Then we see that the functional G(Ψ) is bounded in the norm Ψ1 on this subspace as well, and therefore it can be represented in the form of an inner product [U, Ψ]1 = G(Ψ), where U is an element of the above subspace. Hence, 1 on the subspace of vector functions Ψ ∈ Hper ξ1 (Q) such that Ψ = 0, the integral identity holds for some vector function U from this subspace. Let 1 Ψ(ξ) ∈ Hper ξ1 (Q). Represent Ψ(ξ) in the form Ψ(ξ) = Ψ0 (ξ) + h, where h is a constant h = Ψ and Ψ0 = 0. Then [U, Ψ]1 = [U, Ψ0 ]1 + [U, h]1 = G(Ψ0 ) +

s

i,j=1

Aij

∂h ∂U ∂ξξj ∂ξi

= G(Ψ0 ) = G(Ψ).

1 So the integral identity holds for all functions Ψ(ξ) ∈ Hper ξ1 (Q). This proves the lemma. Remark 2.2.1. A solution U having vanishing average is unique. This follows from the Riesz theorem for G(Ψ). Theorem 2.2.1. h2 > 0.

42

CHAPTER 2. HETEROGENEOUS ROD Proof. Let us obtain a new representation for the matrix s ∂N N1 + A = hN = A 11 1j 2 ∂ξξj j=1

s

=

A1j

j=1

Since

∂ξ1 ∂N N1 + ∂ξξj ∂ξξj

.

s ∂ξ1 Amj = A1j , ∂ξ m m=1

we obtain

s ∂ξ ∂ 1 Amj (N N1 + ξ1 ) . ∂ξm ∂ξξj m,j=1

(2.2.17)

∂ ∂ (N N1 + ξ1 ) . N1 Amj ∂ξξj ∂ξm m,j=1

(2.2.18)

hN 2 =

On the other hand, it follows from the integral identity for problem (2.2.11)– (2.2.14) with the test function N1 that 0=

s

Summing (2.2.17) and (2.2.18), we obtain the following representation for hN 2 : hN 2 =

κ

∂ ∂ (N N1 + ξ1 ) ≥ (N N1 + ξ1 )Amj ∂ξξj ∂ξm m,j=1

s

2 s s 2 ∂ ∂ (N N1 + ξ1 ) / = κ/ > 0. (N N1 + ξ1 ) ≥κ ∂ξξj ∂ξξj j=1 j=1

Theorem 2.2.1 is proved. Equation (2.2.7) then takes the form

= hN 2

∞

P u(∞) − F ψ =

∞

dk ω l M d l ψ d2 ω = 0. + µ hl − F¯ ψ + µk−2 hN k 2 dx1 dxl1 dxk1

(2.2.19)

l=1

k=3

Problem (2.2.19) can be regarded as a homogenized equation of infinite order for ω. The f.a.s. to this problem is sought as the series ω=

∞

µj ωj (x1 ),

(2.2.20)

j=0

where ωj (x1 ) are independent of µ. Substitution of series (2.2.20) into (2.2.19) gives the following recurrent chain of equations for ωj : hN 2

d2 ωj = fj (x1 ), dx21

(2.2.21)

2.2.

43


where fj depend on ωj1 , j1 < j, and their derivatives. Indeed, hN 2

∞

∞

∞

d k ω l M dl ψ d2 ω = + µ hl − F¯ ψ + µk+j−2 hN k 2 dx1 dxl1 dxk1 j=0 l=1

k=3

=

∞ l=0

where So,

hM 0

l−1 l d2 ωl N dl−r+2 ωr Md ψ + h = 0, µl hN + h l 2 l−r+2 dx21 dxl1 dxl−r+2 1 r=0

= −F¯ .

fj (x1 ) = −

j−1

hN j−r+2

r=0

dj−r+2 ωr

dxj−r+2 1

− hM j

dj ψ

dxj1

.

This completes the construction of the f.a.s. for problem (2.2.1), (2.2.2).

2.2.3

Boundary layer corrector.

We construct the f.a.s. to problem (2.2.1)–(2.2.3) in the form u(∞) = uB + u0P + u1P ,

(2.2.22)

where uB is defined by equation (2.2.6) and u0P =

∞

µl Nl0 (ξ)

l=0

u1P =

∞

µl Nl1 (ξ)

l=0

dl ω l+2 0 dl ψ

, + µ Ml (ξ) l dx1 ξ=x/µ dxl1 l=0 ∞

dl ω l+2 1 dl ψ + µ Ml (ξ) l |ξ1 =(x1 −b)/µξ′ =x′ /µ . dx1 dxl1 l=0 ∞

(2.2.23)

Substituting (2.2.22) into (2.2.1)–(2.2.3) and taking into account the fact that uB is as constructed above, we obtain the asymptotic equations ∞

¯ lN µl−2 H

l=0

∞

x dl ω

¯N µl−1 G l

l=0

µ

dxl1

x dl ω

µ dxl1 ∞ l=0

∞ l=0

+

∞

¯M µl H l

l=0

+

∞

¯l ] µl [N

x dl ψ

µ dxl1

¯M µl+1 G l

l=0

∞

− F¯ ψ − Fψ = 0,

x dl ψ

µ dxl1

= 0,

dl ω l+2 ¯ dl ψ + µ [Ml ] l = 0, dx1 dxl1 l=0

x ∈ ∂C Cµ ∩ ∂U Uµ , (2.2.25)

x ∈ Σ, µ

∞ dl ψ x dl ω l+1 ¯ M x ¯N = 0, + µ G µl−1 G l l µ dxl1 µ dxl1

l=0

x ∈ Cµ , (2.2.24)

x ∈ Σ, µ

∞ l l l+2 ¯ ′ d ψ ¯l (0, ξ ′ ) d ω

= 0, + µ M (0, ξ ) µl N

l dxl1 x1 =0 dxl1 x1 =0 l=0 l=0

∞

(2.2.26)

44

CHAPTER 2. HETEROGENEOUS ROD ∞

¯l µl N

l=0

where

b

µ

, ξ′

∞ b dl ψ dl ω

l+2 ¯ , ξ′ = 0, + µ M

l µ dxl1 x1 =b dxl1 x1 =b l=0

(2.2.27)

¯ lN x/µ = HlN x/µ + HlN 0 x/µ + HlN 1 x1 − b/µ, x′ /µ , H N N0 1 ¯N G x/µ + GN x1 − b/µ, x′ /µ , l x/µ = Gl x/µ + Gl l ¯l x/µ = Nl x/µ +N N Nl0 x/µ +N Nl1 x1 − b/µ, x′ /µ , HlN r (ξ) = Lξξ Nlr +T TlN r (ξ), TlN r (ξ) =

s j=1

∂/∂ξξj (Aj1 Nlr−1 ) +

r GN = l

s

m=1

s j=1

A1j ∂N Nlr−1 /∂ξξj + A11 Nlr−2 ;

Amj ∂N Nlr /∂ξξj + Am1 Nlr−1 nm

(2.2.28)

¯M, G ¯M , M ¯ l , T M r , and GM r are obtained by replacing N with M . Here and H l l l l (N,M ) (N,M ) Nl (ξ), Ml (ξ), Hl (ξ), and Gl (ξ) are the 1-periodic functions in ξ1 con(N,M )r (N,M )r structed above and Nlr (ξ), Hl (ξ), and Gl (ξ) stabilize exponentially to zero as ξ1 → ±∞ (+ for r = 0 and − for r = 1). The superscript (N, M ) means that the assertion is true for both N and M . The exponential stabilization to zero implies that |N Nlr (ξ)|dξ, |M Mlr (ξ)|dξ ≤ c1 e−c2 |σ| as σ → ±∞ (σ,σ +1)×β

(σ,σ +1)×β

(as above, + for r = 0 and − for r = 1). Here c1 and c2 are positive constants independent of σ. We arrange Nlr and Mlr in ascending order with respect to the index l and (N,M )r note that Tl is defined by the functions Nlr1 (or Mlr1 ) for l1 < l. We require (N,M )r ¯ ¯ l (0, ξ ′ ), N ¯l (b/ε, ξ ′ ), and M ¯ l (b/ε, ξ ′ ) satisfy that H , Nl (0, ξ ′ ), M l

(N,M )r

Hl

= 0,

(N,M )r

Gl

= 0,

¯ l (0, ξ ′ ) = hM 0 + O(e−c/µ ), M l

0 ¯l (0, ξ ′ ) = hN N + O(e−c/µ ), l

1 ¯ l b , ξ ′ = hN + O(e−c/µ ), N l µ

¯ l b , ξ ′ = hM 1 + O(e−c/µ ), M l µ

c > 0.

(N,M )r

(2.2.29)

The constants hl are to be defined from the conditions of the existence of functions Nlr and Mlr exponentially stabilizing to zero. We construct the functions Nlr , Mlr satisfying equations (2.2.29) by induction on l. Assume that Nlr , Mlr = 0 for l < 0. Let all Nlr1 , Mlr1 be constructed for 0 such that there exists a solution, exponentially l1 < l. We choose a constant hN l stabilizing to zero, to the problem Lξξ Nl0 + TlN 0 = 0,

ξ1 > 0,

ξ ′ ∈ β,

2.2.

STEADY-STATE CONDUCTIVITY OF A ROD s ∂N Nl0 =− Am1 Nl−1 nm , ∂ννξ m=2

45

ξ ′ ∈ ∂β,

0 Nl0 (0, ξ ′ ) = −N Nl (0, ξ ′ ) + hN l ,

(2.2.30)

Figure 2.2.3. A boundary layer problem. 1 and choose a constant hN such that there exists a solution exponentially l stabilizing to zero to the problem Lξξ Nl1 + TlN 1 = 0,

ξ1 < 0,

s ∂N Nl1 =− Am1 Nl1−1 nm , ∂ννξ m=2

ξ ′ ∈ β,

ξ ′ ∈ ∂β,

1 Nl1 (0, ξ ′ ) = −N Nl (0, ξ ′ ) + hN l .

Mlr

(2.2.31)

r hM l

The functions and constants are defined similarly (by replacing N with M in equations (2.2.30) and (2.2.31)). The following theorem provides the possibility of such a choice of the con(N,M )r . stants hl Theorem 2.2.2. Let Aij (ξ) satisfy the above conditions, let Fj (ξ) ∈ L2 ([0, + ∞)× β), j = 0, 1, . . . , s, satisfy |F Fj (ξ)|dξ ≤ c1 e−c2 σ , σ > 0, (2.2.32) (σ,σ +1)×β

′

and let u0 (ξ ) ∈ H the problem

1/2

({0} × β) be given. Then there exists a solution u(ξ) to

Lξξ u = F0 (ξ) +

s ∂F Fj j=1

∂ξξj

,

ξ1 > 0,

ξ ′ ∈ β,

(2.2.33)

46

CHAPTER 2. HETEROGENEOUS ROD s

∂u Fj nj , ξ ′ ∈ ∂β, = ∂ννξ j=2

u ξ =0 = u0 (ξ ′ )

(2.2.34) (2.2.35)

1

such that for some constant h, the following inequality holds |u − h|dξ ≤ c1 e−c2 σ , c1 , c2 > 0, σ > 0.

(2.2.36)

(σ,σ +1)×β

Remark 2.2.2. Let us explain some terms from Theorem 2.2.2 (see [89],[90], [129],[130]). The space H 1/2 ({0} × β) is thought of as the space of traces V (ξ ′ ) on the set {ξ1 = 0, ξ ′ ∈ β} of functions from H 1 ((0, 1) × β) equipped with the norm s

(∂v/∂xj )2 +|v|2 dξ, v ∈ H 1 ((0, 1)×β), v ξ =0 = V (ξ ′ ) . V 21/2 = inf v

1

(0,1)×β j=1

A solution to problem (2.2.33)–(2.2.35) is sought in the form of a function u(ξ) ∈ H 1 loc whose trace is equal to u0 (ξ ′ ) for ξ1 = 0. For any function Ψ(ξ) ∈ C ∞ with support in the cylinder (δ, D) × β, δ > 0, the function u(ξ) must satisfy the integral identity −

s

i,j=1

(0,+∞)×β

Aij

∂u ∂Ψ dξ = ∂ξξj ∂ξi

(0,+∞)×β

F0 Ψdξ −

s j=1

(0,+∞)×β

Fj

∂Ψ dξ. ∂ξξj

Theorem 2.2.2 follows directly from Theorems 4 and 5 in [130]. Their proofs are ”corrected” for the right-hand side having specific representation (2.2.33) and (2.2.34) and the momenta in Theorem 5 are chosen so that they are equal to zero at infinity . 0 To determine the hN from (2.2.30), it suffices to first solve the problem l 0 + T N 0 = 0, Lξξ N l l

ξ ′ ∈ β,

ξ1 > 0,

s 0 ∂N l =− nm Am1 Nl0−1 , ∂ννξ m=2

ξ ′ ∈ ∂β,

l0 (0, ξ ′ ) = −N N Nl (0, ξ ′ ).

(2.2.37)

0 stabilizes to a constant h. If one subtracts h According to Theorem 2.2.2, N l 0 0 (ξ) − h will stabilize to zero and satisfy from Nl (ξ), then the difference Nl0 = N l N0 r (2.2.30) for hl = −h. Problem (2.2.31) and the problems for Mlr and hM can l be solved in the same manner.

2.2.

47


0 Figure 2.2.4. How to determine hN l . It follows from Theorem 2.2.2 that the energy e of the columns Nlr and Mlr also exponentially stabilizes to zero. After the successive determination of all Nlr and Mlr by induction (note that r N0 = 0) we obtain a homogenized problem of infinite order for the function ω(x1 ). This problem is given by equation (2.2.17) with the boundary conditions

ω+

∞ l=1

r µl hN l

dl ω l+2 M r dl ψ

= 0, + µ hl

dxl1 x1 =rb dxl1 l=0 ∞

r = 0, 1,

(2.2.38)

48


r r It is significant that hN and hN are equal to 1. 0 1 The f.a.s. to problem (2.2.19), (2.2.38) and (2.2.39) is represented in the form of the series (2.2.20) and, being substituted into (2.2.19), (2.2.38) and (2.2.39), leads to the recurrent chain of equations (2.2.21) for ωj with the boundary conditions

ωj x1 =rb = gjr , r = 0, 1, (2.2.39)

Here gjr depend on ωj1 with j1 < j, and on the derivatives of these functions. Indeed, ∞

µj ωj +

j=0

=

r µl+j hN l

l=1 j=0

∞

µj ωj +

j=0

so,

∞ ∞

j−1

r hN j−p

p=0

dl ωj l+2 M r dl ψ |x =rb = + µ hl dxl1 1 dxl1 l=0 ∞

dj−p ωp

dx1j−p

r + hM j−2

dj−2 ψ dxj−2 1

|x1 =rb = 0,

j−1 j−2 j−p ψ ωp r d r d − hM |x1 =rb . hN gjr = − j−2 j−p j−p j−2 dx dx 1 1 p=0

The chain of problems (2.2.21),(2.2.39) is consecutively solvable. This completes the construction of the f.a.s. for problem (2.2.1)-(2.2.3).

2.2.4

The justification of the asymptotic expansion

Here we prove the following main theorem. Theorem 2.2.3. Let K ∈ {0, 1, 2, ...}. Denote χ a function from C (K+2) ([0, b]) such that χ(t) = 1 when t ∈ [0, b/3], χ(t) = 0 when t ∈ [2b/3, b]. Consider a function (K)

0(K)

u(K) (x) = uB (x) + uP

1(K)

(x)χ(x1 ) + uP

(x)χ(b − x1 ) + ρ(x1 ),

where (K)

uB

=

K+1

µl Nl

l=0

0(K)

uP

=

=

K+1

K+1

x dl ψ(x ) x dl ω (K) (x ) K−1 1 1 , + µl+2 Ml l l µ µ dx dx1 1 l=0

µl Nl0 (ξ)

l=0

1(K)

uP

l=0

µl Nl1 (ξ)

K−1 dl ψ

dl ω (K) , + µl+2 Ml0 (ξ) l l dx1 ξ=x/µ dx1 l=0

K−1 dl ψ dl ω (K) + µl+2 Ml1 (ξ) l |ξ1 =(x1 −b)/µξ′ =x′ /µ , l dx1 dx1 l=0

ω (K) =

K j=0

µj ωj (x1 );

2.2.


49

ρ(x1 ) = (1 − x1 /b)q0 + (x1 /b)q1 ,

qr =

2K+1

l=K+1

µl

r hN p

j+p=l,0≤j≤K,0≤p≤K+1

K−1 ψ d p ωj K+1 M r d |x1 =rb , r = 0, 1, +µ h p K−1 K−1 dx1 dx1

Nl , Nl0 , Nl1 , Ml , Ml0 , Ml1 , ωj are defined in subsection 2.3. Then the estimate holds u(K) − uH 1 (Cµ ) = O(µK ) mes Cµ .

Proof. Let us estimate the discrepancy functional

I(φ) =

s

Amj

Cµ m,j=1

x ∂(u − u(K) ) ∂φ dx, ∂xj ∂xm µ

where φ ∈ H 1 (C Cµ ), φ = 0 for x1 = 0 or b. 1. At the first stage represent it in the form I(φ) = F ψφ dx − JB (φ) − J0 (φ) − J1 (φ) − J2 (φ), Cµ

where

JB (φ) =

s

Amj

Cµ m,j=1

J0 (φ) =

s

Amj

Cµ m,j=1

J1 (φ) =

s

Amj

Cµ m,j=1

J2 (φ) =

x ∂(u0(K) (x)χ(x )) ∂φ 1 P dx, ∂xm ∂xj µ

x ∂(u1(K) (x)χ(b − x )) ∂φ 1 P dx, ∂xm ∂xj µ

s

Amj

x ∂ρ(x ) ∂φ 1 dx = µ ∂xj ∂xm

s

Am1

Cµ m,j=1

q1 − q 0 = b

so,

x ∂u(K) ∂φ B dx, µ ∂xj ∂xm

Cµ m=1

x ∂φ dx; µ ∂xm

|J J2 (φ)| ≤ c0 µK+1 φH 1 (Cµ )

mes Cµ .

50


The boundary layer functions Nlr , Mlr decay exponentially as |ξ1 | → +∞, so (2.2.3) implies the following estimates

|

|

x ∂(u0(K) (x)χ(x )) ∂φ 1 P dx| ≤ c1 e−c2 /µ φH 1 (Cµ ) , ∂xm ∂xj µ

s

Amj

s

Amj

Cµ ∩{x1 ≥b/3} m,j=1

Cµ ∩{x1 ≤2b/3} m,j=1

x ∂(u1(K) (x)χ(b − x )) ∂φ 1 P dx| ≤ c1 e−c2 /µ φH 1 (Cµ ) ∂xm ∂xj µ

with some constants c1 , c2 > 0 independent of µ. So I(φ) = F ψφ dx − JB (φ) − Jˆ0 (φ) − Jˆ1 (φ) + Cµ

+ (Jˆ0 (φ) − J0 (φ)) + (Jˆ1 (φ) − J1 (φ)) − J2 (φ), where Jˆr (φ) =

s

Amj

Cµ m,j=1

and

x ∂ur(K) (x) ∂φ P dx, r = 1, 2, ∂xm ∂xj µ

|Jˆr (φ) − Jr (φ)| ≤ c1 e−c2 /µ φH 1 (Cµ ) . Consider now JB (φ) = =

K+1 s

Cµ

l=0 m=1

+Am1

+

K−1

x

s

µ

l=0 m=1

µl−1

Amj

j=1

Nl−1

Cµ

s

x ∂N Nl (ξ) |ξ=x/µ + ∂ξξj µ

x dl ω (K) (x ) ∂φ 1 dx + ∂xm µ dxl1

µl+1

s j=1

Amj

x ∂M Ml (ξ) |ξ=x/µ + ∂ξξj µ

x dl ψ(x ) ∂φ 1 dx + Ml−1 +Am1 µ µ dxl1 ∂xm s x dK+2 ω (K) (x1 ) ∂φ dx + NK +1 + µK+1 Am1 ∂xm µ dxK+2 1 m=1 Cµ s x dK+2 ψ(x1 ) ∂φ dx. MK −1 + µK+1 Am1 ∂xm µ dxK+2 1 m=1 Cµ x

2.2.

51


Denote s

Amj (ξ)

∂N Nl (ξ) + Am1 (ξ)N Nl−1 (ξ), ∂ξξj

s

Amj (ξ)

∂M Ml (ξ) + Am1 (ξ)M Ml−1 (ξ), ∂ξξj

ABN ml (ξ) =

j=1

ABM ml (ξ) =

j=1

∆N 1 (φ)

s

=

µK+1 Am1 (x/µ)N NK +1

Cµ

m=1

s

∆M 1 (φ) =

Cµ

m=1

dK+2 ω (K) (x1 ) ∂φ dx, ∂xm dxK+2 1

µK+1 Am1 (x/µ)M MK −1

dK+2 ψ(x1 ) ∂φ dx. ∂xm dxK+2 1

We have, M K+1 |∆N φH 1 (Cµ ) 1 (φ)| , |∆1 (φ)| ≤ c3 µ

with a constant c3 > 0 independent of µ. We get

mes Cµ ,

JB (φ) = K+1

=

s

Cµ

l=0 m=1

K−1

+

s

µl−1 ABN ml (x/µ)

Cµ

l=0 m=1

dl ω (K) (x1 ) ∂φ dx + ∂xm dxl1

µl+1 ABM ml (x/µ)

dl ψ(x1 ) ∂φ dx + dxl1 ∂xm

M + ∆N 1 (φ) + ∆1 (φ).

2. At the second stage consider the sum N J˜B (φ) =

=

K+1

s

l=0 m=1

=

Cµ


K+1

s


−

K+1

µl−1 ABN 1l (x/µ)φ

Cµ

l=0 m=1

l=0

Cµ

dl ω (K) (x1 ) ∂φ dx = ∂xm dxl1

∂ dl ω (K) (x1 ) dx − φ ∂xm dxl1

dl+1 ω (K) (x1 ) dx = dxl+1 1

52


=

K+1

s

µ−1 Cµ

l=0 m=1

−

where

µs+l−2 ABN ml (ξ)

∂ dl ω (K) (x1 ) |x=µξ dξ − φ ∂ξm dxl1

K+1

dl+1 ω (K) (x ) 1 |x=µξ dξ, µs+l−1 ABN (ξ) φ 1l dxl+1 µ−1 Cµ 1

l=0

µ−1 Cµ = {ξ ∈ IRs | µξ ∈ Cµ } = (0, b/µ) × β. Finally

N J˜B (φ) =

K+1

µs+l−2

µ−1 Cµ

l=0

s

ABN ml (ξ)

m=1

∂ dl ω (K) (x1 ) |x=µξ − (φ ∂ξm dxl1

dl ω (K) (x ) 1 − ABN |x=µξ dξ − 1,l−1 (ξ) φ dxl1 − µs+K

dK+2 ω (K) (x ) 1 |x=µξ dξ, ABN 1,K+1 (ξ) φ dxK+2 µ−1 Cµ 1

where ABN ml = 0 when l < 0 (by convention). Denote s+K ∆N 2 (φ) = µ

dK+2 ω (K) (x ) 1 |x=µξ dξ. ABN 1,K+1 (ξ) φ dxK+2 µ−1 Cµ 1

3. The variational formulation for problem (2.2.11)-(2.2.13) gives: 1 ˜ ∀φ(ξ) ∈ Hper

s

m=1

ABN ml (ξ)

ξ1 (Q),

∂ φ˜ ˜ ˜ − ABN (ξ) φ(ξ) = hN 1,l−1 l φ(ξ). ∂ξm

1 Then this identity holds, in particular, for any function φ˜ ∈ Hper ξ1 (Q), vanishing when ξ1 = 0 (and therefore when ξ1 = 1). So, this identity holds for any function φ˜ ∈ H 1 (Q), vanishing when ξ1 = 0 and when ξ1 = 1, as well as for any function φ˜ ∈ H 1 ((a, a + 1) × β), vanishing when ξ1 = a and when ξ1 = a + 1; in the last case the average is defined as (mes β)−1 (a,a+1)×β dξ. Let φ˜ be a function from H 1 (µ−1 Cµ ) vanishing when ξ1 = 0 and when ξ1 = b/µ. We represent it in the form

˜ ˜ ˜ φ(ξ) = φ(ξ) sin2 (πξ1 ) + φ(ξ) cos2 (πξ1 ),

2.2.

53


˜ sin2 (πξ1 ) vanishes when ξ1 = i, i ∈ {0, 1, ..., b/µ} and φ(ξ) ˜ cos2 (πξ1 ) where φ(ξ) vanishes when ξ1 = i − 1/2, i ∈ {1, ..., b/µ} as well as when ξ1 = 0 or ξ1 = b/µ. Therefore, by simple addition the identity can be generalized as

s

ABN ml (ξ)

µ−1 Cµ m=1

∂ φ˜ ˜ − ABN 1,l−1 (ξ)φ(ξ) dξ = ∂ξm

µ−1 Cµ

˜ hN l φ(ξ) dξ

˜ for any φ(ξ) ∈ H 1 (µ−1 Cµ), vanishingwhen ξ1 = 0 or ξ1 = b/µ; for instance, it l (K) ˜ remains valid for φ(ξ) = φ d ω l(x1 ) |x=µξ . dx1

Thus,

N J˜B (φ) =

K+1

dl ω (K) (x ) 1 |x=µξ dξ + ∆N φ µs+l−2 hN 2 (φ) = l l −1 dx µ Cµ 1

l=0

K+1

=

Cµ

l=0

where

µl−2 hN l φ(x)

dl ω (K) dx + ∆N 2 (φ), dxl1

(N )

K |∆N 2 (φ)| ≤ c4 µ φH 1 (Cµ )

(N )

with a constant c4 > 0 independent of µ. 4. The same reasoning for M J˜B (φ) =

K−1

s

F˜ φ +

l=0 m=1

Cµ

µl+1 ABM ml (x/µ)

mes Cµ ,

dl ψ(x1 ) ∂φ dx dxl1 ∂xm

gives M J˜B (φ) =

Cµ

where

K−1

µl hM l φ(x)

l=1

(M ) K

|∆M 2 (φ)| ≤ c4

dl ψ dx + ∆M 2 (φ), dxl1

µ φH 1 (Cµ )

(M )

mes Cµ ,

with a constant c4 > 0 independent of µ. 0 5. Replacing Nl by Nl0 in the expressions ABN ml and replacing Ml by Ml in BM 0N 0M the expressions Aml we define Aml and Aml respectively. Applying the same reasoning as at stages 2 and 3 we obtain the relation

Jˆ0 (φ) =

K+1 l=0

µ−1 Cµ

µs+l−2

s

m=1

A0N ml (ξ)

∂ dl ω (K) (x1 ) |x=µξ − (φ ∂ξm dxl1

dl ω (K) (x ) 1 |x=µξ dξ − − A0N 1,l−1 (ξ) φ dxl1

54


s+K

−µ

+

K−1

dK+2 ω (K) (x ) 1 |x=µξ dξ + A0N 1,K+1 (ξ) φ K+2 −1 dx1 µ Cµ

µs+l

µ−1 Cµ

l=0

s

A0M ml (ξ)

m=1

∂ dl ω (K) (x1 ) |x=µξ − (φ ∂ξm dxl1

dl ω (K) (x ) 1 − A0M (ξ) φ |x=µξ dξ − 1,l−1 dxl1 dK ψ(x ) 1 |x=µξ dξ. − µs+K A0M (ξ) φ 1,K+1 dxK µ−1 Cµ 1

Variational formulation for problem (2.2.30) gives

s

µ−1 Cµ m=1

A0N ml (ξ)

∂ φ˜ ˜ − A0N 1,l−1 (ξ)φ(ξ) dξ = 0 ∂ξm

for any function φ˜ ∈ H 1 (µ−1 Cµ ) vanishing when ξ1 = 0 or ξ1 = b/µ. Analogously,

s

µ−1 Cµ m=1

A0M ml (ξ)

∂ φ˜ ˜ − A0M 1,l−1 (ξ)φ(ξ) dξ = 0 ∂ξm

for any function φ˜ ∈ H 1 (µ−1 Cµ ) vanishing when ξ1 = 0 or ξ1 = b/µ. So, |Jˆ0 (φ)| ≤ c5 µK φH 1 (Cµ ) mes Cµ ,

with a constant c5 > 0 independent of µ. Similarly, |Jˆ1 (φ)| ≤ c6 µK φH 1 (Cµ ) mes Cµ , with a constant c6 > 0 independent of µ. Thus, I(φ) =

Cµ

−F˜ ψ −

K−1 l=1

Fψ −

µl hM l

K+1 l=0

µl−2 hN l

dl ω (K) − dxl1

dl ψ φ(x) dx + ∆3 (φ), dxl1

where ∆3 (φ) is a linear functional of φ such that |∆3 (φ)| ≤ c7 µK φH 1 (Cµ ) mes Cµ ,

with a constant c7 > 0 independent of µ. Note that F ψ − F˜ ψ = F ψ. Consider the expression

2.2.

55


B(x1 ) = F ψ −

−F˜ ψ −

K+1

µl−2 hN l

l=0

K−1

µl hM l

l=1

dl ω (K) − dxl1

dl ψ = dxl1

K+1 K−1 d2 ω (K) l (K) l l−2 N d ω l Md ψ ¯ψ + . − F µ h − µ h = − hN l l 2 dx21 dxl1 dxl1 l=3

K

(K)

l=1

j

Substitute ω (x1 ) = j=0 µ ωj (x1 ) and remind (2.2.21). Then there exist a constant c8 independent of µ such that

|B(x1 )| ≤ c8 µK . Thus |I(φ)| ≤ |

Cµ

B(x1 )φ dx| + |∆3 (φ)| ≤

≤ c8 µK φH 1 (Cµ )

mes Cµ + |∆3 (φ)| ≤

≤ (c8 + c7 )µK φH 1 (Cµ ) mes Cµ ≤ ≤ c9 µK φH 1 (Cµ ) mes Cµ ,

with a constant c9 > 0 independent of µ. Cµ ) and it vanishes when x1 = 0 or x1 = b. On the other hand, u(K) ∈ H 1 (C Indeed, (K)

u(K) |x1 =0 = (uB =

K+1 l=0

+

K−1 l=0

=

K+1 l=0

0(K)

+ uP

)|x1 =0 + q0 =

x x dl ω (K) (x1 ) |x1 =0 + |x1 =0 µl Nl + Nl0 µ µ dxl1

x x dl ψ(x1 ) |x1 =0 + q0 = |x1 =0 + Ml0 µl+2 Ml µ µ dxl1

0 µl hN l

K−1 l dl ω (K) l+2 M 0 d ψ |x =0 + q0 = 0 | + µ h x =0 l 1 dxl1 1 dxl1 l=0

due to relations (2.2.39). Similarly, u(K) |x1 =b = 0. Taking φ = u − u(K) we obtain the inequality κu − u(K) 2H 1 (Cµ ) ≤ I(u − u(K) ) =

56


=

s

Amj

Cµ m,j=1

x ∂(u − u(K) ) ∂(u − u(K) )

∂xm

∂xj

µ

≤ c10 µK u − u(K) H 1 (Cµ )

dx ≤

mes Cµ ,

(2.2.40)

with a constant c10 > 0 independent of µ. So, u − u(K) H 1 (Cµ ) ≤ c10 /κµK mes Cµ ;

this completes the proof of the theorem. Corollary 2.2.1. The estimate holds (K−1)

u − uB

0(K−1)

− uP

1(K−1)

− uP

H 1 (Cµ ) = O(µK ) mes Cµ , K = 1, 2, ....

1(K) Indeed, the H 1 −norm of the differences u0(K) − u0(K) − χ(x1 ) and u 1 K u χ(b − x1 ) and the H −norm of ρ are of order O(µ ) mes Cµ . Moreover, (K) (K−1) 0(K) 0(K−1) 1(K) 1(K−1) uB − uB H 1 (Cµ ) , uP − uP H 1 (Cµ ) , uP − uP H 1 (Cµ ) = K O(µ ) mes Cµ . 1(K)

(K−1)

So when one replaces in the estimate of the theorem u(K) by uB + 0(K−1) 1(K−1) uP + uP , the order of this estimate does not change. It proves the corollary. (K−1) 0(K−1) 1(K−1) Remark 2.2.3. The function uB + uP + uP does not vanish K at the ends of the rod; however, it has the order O(µ ) there. Remark 2.2.4. In the above proof we have estimated some linear functionM ˆ ˆ als: ∆N j (φ), ∆j (φ), j = 1, 2, ∆3 (φ), J2 (φ), J0 (φ), J1 (φ), as well as the differ ences Jˆr (φ) − Jr (φ), r = 1, 2, by the upper born cµK φH 1 (Cµ ) mes Cµ (for sufficiently small µ). Finally, we have obtained the same upper born for the functional I(φ). One can easily check that all these functionals can be presented in a form of the H 1 −inner product

Cµ

Ψ0,µ (x)φ dx +

s

m=1

Ψm,µ (x)

Cµ

∂φ dx, ∂xm

(2.2.41)

where functions Ψ0,µ , Ψm,µ , m = 1, ..., s, are estimated as Ψ0,µ L2 (Cµ ) , Ψm,µ L2 (Cµ ) ≤ cµK

with a constant c > 0 independent of µ.

2.3

mes Cµ .

(2.2.42)

Steady state elasticity equation in a rod

Constructions of section 2.2.3 repeat some ideas of section 2.2.2 but the elasticity system of equations is essentially more complex than the conductivity equation.

2.3. STEADY STATE ELASTICITY EQUATION IN A ROD

2.3.1

57

Formulation of the problem

We consider the elasticity system of equations (see [55]) Pu ≡

s ∂ x ∂u = Fµ (x), Aij µ ∂xj ∂xi i,j=1

x ∈ Cµ

(2.3.1)

with the boundary conditions s x ∂u ∂u ni = 0, ≡ Aij µ ∂xj ∂ν i,j=1

u = 0,

x1 = 0

x ∈ ∂U Uµ ,

or x1 = b.

(2.3.2) (2.3.3)

Here (n1 , . . . , ns ) is an exterior vector normal to the boundary Uµ and Aij (ξ) are (s × s)-matrices whose elements akl ij (ξ) satisfy the conditions kj lk akl ij (ξ) = ail (ξ) = aji (ξ),

and there exists a constant κ > 0 such that for any symmetric matrix ηik , for any ξ ∈ IRs the following inequality holds: s

i,j,k,l=1

k l akl ij (ξ)ηi ηj ≥ κ

s

(ηik )2 ,

κ > 0.

i,k=1

It is assumed that the functions Aij (ξ) are 1-periodic in ξ1 , infinitely differentiable everywhere outside of a set Σ (consisting of smooth nonintersecting surfaces Σl ) up to Σ; we assume that there is only a finite number of surfaces ∞ Σl that intersect with the cylinder [0, b] × β; Σ ∩ ([0, 1] × ∂β) = ∅, akl ij ∈ C kl and on Σ the coefficients have jump discontinuities. If aij depend on ξ1 , we require that b/µ be an integer. The right-hand side Fµ and the unknown u are s-dimensional vector-functions, and Fµ has the following structure: x′ x Ψ(x1 ); F Fµ = Φ µ µ

here Φ(ξ ′ ) is the matrix of rigid displacements; for s = 2, Φ(ξ ′ ) = I is the 2 × 2 identity matrix, and for s = 3 we have ⎞ ⎛ 1 0 0 0 (ξ 2 + ξ 2 ) dξ dξ −1/2 2 3 2 3 ′ ⎝ ⎠ 0 1 0 −aξ3 , a = Φ(ξ ) = ; mes β β 0 0 1 aξ2 F (ξ) is a d × d square matrix with d = 2 if s = 2 and d = 4 if s = 3; ψ(x1 ) is a d-dimensional vector function.

58


Figure 2.3.1. Right hand side force structure It is assumed that F (ξ) is a 1-periodic matrix function of ξ1 , which has the same smoothness as the coefficients (see above) and that ψ ∈ C ∞ ([0, b]). On the interfaces of discontinuity of the coefficients, the natural conjugation conditions are assumed: [u] = 0;

s

i,j=1


(2.3.4)

where (n1 , . . . , ns ) is the normal vector to the interface of discontinuity, [·] is the jump of a function on this interface. For s = 3, problem (2.3.1)–(2.3.3) models the stress-strain state of a rod of length b with cross-section βµ = {x′ ∈ R2 , x′ /µ ∈ β}; the ends of the rod are clamped and the lateral surface is free. The rod has an inhomogeneous structure: the elements of the elasticity tensor akl ij are µ-periodic functions of the longitudinal coordinate x1 , they also depend on the transverse coordinates x′ . The right-hand side (mass forces) has a tensile-pressing part (the first component of the vector F (x/µ)ψ(x1 )), bending parts (the second and third components of F (x/µ)ψ(x1 )) and a torsional part (the fourth component of F (x/µ)ψ(x1 )), each part is the product of the fast-oscillating function F (x/µ) by the slowly varying factor ψ(x1 ). For s = 2, the problem models a 2-D stress-strain state of the plate of length b and thickness µ. The right-hand side here has both tensile-pressing and bending components. We construct formal asymptotic solutions (f.a.s.) to problem (2.3.1)–(2.3.3) in two stages, as in section 1. At the first stage we construct f.a.s. to equation (2.3.1) with condition (2.3.2), using an analog of Bakhvalov’s ansatz; at


59

the second stage we construct the boundary layer corrector. For the reader’s convenience, we first establish a formalism for each stage and then justify the asymptotic expansion, i.e., prove different assertions stated in constructing f.a.s. Furthermore, we prove that f.a.s. is an asymptotic expansion of an exact solution to the problem and establish an error estimate for the partial sum of the series of f.a.s. This justification is developed for the rods with the symmetry of a periodic cell with respect to two co-ordinate planes (Condition A in section 2.3.2).

2.3.2

Inner expansion

Let us represent the right-hand side F as the sum F = F¯ + F,

(2.3.5)

where F¯ = Φ∗ ΦF , Φ∗ ΦF = 0. From now on in this section F (ξ, x) denotes the average F (ξ, x)dξ . mes β (0,1)×β

We seek f.a.s. in the form of an analog of Bakhvalov’s ansatz, u(∞) =

∞

µl Nl

l=0

∞ x dl ψ(x ) x dl ω(x ) 1 1 , + µl+2 Ml l l µ µ dx dx1 1 l=0

(2.3.6)

where Nl , Ml are 1-periodic in ξ1 matrix functions, ω(x1 ) is a d - vector function. Substituting series (2.3.6) into (2.3.1), (2.3.2), and (2.3.4) and collecting terms with like powers of µ yields P u(∞) − ΦF ψ =

∞

µl−2 HlN (ξ)

l=0

∞

dl ω l M dl ψ + µ H (ξ) l dxl1 dxl1 l=0

where

− ΦF ψ − ΦFψ,

(2.3.7)

HlN (ξ) = Lξξ Nl + TlN (ξ), s ∂ ∂ , Ajm Lξξ = ∂ξm ∂ξξj j,m=1

TlN (ξ) =

s s ∂N Nl−1 ∂ + A11 Nl−2 ; (Aj1 Nl−1 ) + A1j ∂ξξj ∂ξξj j=1 j=1

∞

∞

l=0

l=0

dl ψ dl ω l+1 M ∂u(∞) , + µ G (ξ) = µl−1 GN l l (ξ) ∂ν dxl1 dxl1

GN l =

s s

m=1

j=1

Amj

∂N Nl + Am1 Nl−1 nm ; ∂ξξj

(2.3.8)

(2.3.9)

(2.3.10)

60


HlM , TlM , GM l are obtained by replacing N with M . Here, as in section 2.2, Nl = 0 whenever l < 0. We require that, as in the procedure from section 2.2, HlN (ξ) = Φ(ξ ′ )hN l ,

HlM (ξ) = Φ(ξ ′ )hM l ,

l ≥ 0,

l > 0,

( ), H0M (ξ) = Φ(ξ ′ )F(ξ

M where hN are constant d × d matrices. Moreover, we require that l and hl N M Gl (ξ) = 0, Gl (ξ) = 0 on the surface (0, 1) × ∂β and that [N ] = 0, [GN l ] = 0; [M ] = 0, [GM ] = 0 on the discontinuity interfaces Σ of the coefficients akl ij (ξ). l We obtain the following recurrent chain of problems for Nl , Ml :

Lξξ Nl = −T TlN (ξ) + ΦhN l , ∂/∂ννξ Nl = −

[N Nl ] Σ = 0,

s

m=2

Am1 Nl−1 nm ,

ξ ∈ (0, 1) × β,

(2.3.11)

ξ ∈ (0, 1) × ∂β,

(2.3.12)

s

∂N Nl /∂ννξ Σ = − Am1 Nl−1 nm Σ .

(2.3.13)

m=1

Here, as in section 2.2, hN l are chosen from the solvability conditions for (2.3.11), (2.3.12), (2.3.13) (see further Lemma 2.3.1): Φ∗ (−T TlN + ΦhN l ) = −

s

m=2

Φ∗ Am1 Nl−1 nm (0,1)×∂β −

s

m=1

Φ∗ [Am1 Nl−1 nm ]Σ ,

where for the (s − 1)-dimensional hyper-surface Γ we have dξ , ·Γ = mes β Γ

i.e.,

s ∂N Nl−1 ∗ + A11 Nl−2 . hN = Φ A1j l ∂ξξj j=1

(2.3.14)

Ml are the solutions of the same problems with Nl replaced by Ml . However, M0 is the solution to the problem ∂M

M0

∂M M0

= 0, [M M0 ] Σ = 0, Lξξ M0 = ΦF,

= 0.

′ ∂ννξ Σ ∂ννξ ξ ∈∂β (2.3.15) Thus, the algorithm for constructing the matrices Nl and Ml is inductive. Suppose that Nl = 0, Ml = 0 for l < 0, N0 = Φ, and M0 is the solution to problem (2.3.15). If l > 0, then Nl and Ml are the solutions to problems (2.3.11)–(2.3.14). The right-hand sides of these problems contain Nm and Mm with indices m < l, which permits one to define them successively. We have

hM 0 = 0,

hN 0 = 0,

hN 1 = 0,

s ∂N N1 ∗ +A11 Φ . (2.3.16) hN = Φ A1j 2 ∂ξξj j=1


61

Here a 1-periodic solution Nl to problem (2.3.11)-(2.3.13) is understood in s×s 1 the same sense as in section 2.2. Let Hper be space of matrix valued ξ1 (Q) 1 functions s × s with the components from Hper ξ1 (Q). s×s ¯ ∈ H1 A solutionU to the problem per ξ1 (Q)

¯ = F¯0 (ξ) + Lξξ U

s ∂ F¯j j=1

s ¯ ∂U = F¯j nj , ∂ννξ j=2

∂ξξj

,

ξ ∈ IR × β,

ξ ∈ IR × ∂β,

is understood as an s×s matrix valued function U satisfying the integral identity

s s ¯ ∂U ∂Ψ∗ ¯ ∂Ψ∗ Fj Aij = Ψ∗ F¯0 − − ∂ξξj ∂ξi ∂ξξj j=1 i,j=1

s×s 1 for any s × s matrix valued function Ψ ∈ Hper . Here the right-hand ξ1 (Q) ¯ ¯ ¯ sides F0 , F1 , . . . , Fs are s×s matrix valued functions such that their components belong to L2 ((−A, A) × β) for each A and are 1-periodic in ξ1 . This problem can be easily reduced to s independent problems for columns of ¯ with right hand sides equal to corresponding columns of the matrices matrix U ¯ we have the F0 , F1 , . . . , Fs . In this case for each column U of the matrix U following variational formulation: 1 find a column U ∈ Hper the integral identity holds

ξ1 (Q)

s

1 such that for any column Ψ ∈ Hper

ξ1 (Q)

s s ∂Ψ ∂U ∂Ψ . = (F F0 , Ψ) − Fj , − Aij , ∂ξξj ∂ξξj ∂ξi j=1 i,j=1 Here the right-hand sides F0 , F1 , . . . , Fs are vector valued functions (corresponding columns of the matrices F¯0 , F¯1 , . . . , F¯s ) such that their components belong to L2 ((−A, A) × β) for each A and are 1-periodic in ξ1 . Lemma 2.3.1. Assume that the right-hand sides F1 , . . . , Fs are 1-periodic s-dimensional vector functions of ξ1 with components satisfying Fij = Fji . The problem has a solution if and only if Φ∗ F0 = 0. Proof. The proof similar to the proof of lemma 2.2.1 and it is also based on the Riesz representation theorem for a bounded linear functional on a Hilbert space.

s

,

62


Namely, the right-hand side of the integral identity (2.2.82) represents the linear

functional G(Ψ) that is bounded in the norm of H 1 (Q) |G(Ψ)| ≤

s j=0

s

by the inequality

F Fj (L2 (Q))s Ψ(H 1 (Q))s ,

s On the other hand, for vector functions Ψ ∈ H 1 (Q) that satisfy Φ∗ Ψ = 0

and the Korn inequality e(Ψ) ≥ c1 Ψ2(H 1 (Q))s , c1 > 0, holds (it can be proved by repeating the proof of [55](see also [129]), Theorem 12.11 and the remark to this theorem literally). Here we write e(Ψ) =

Since

s i,j=1

the norm

2 s ∂Ψj ∂Ψi . + ∂ξi ∂ξξj i,j=1

∂Ψ ∂Ψ , Aij ∂ξξj ∂ξi

≥ κe(Ψ),

s ∂Ψ ∂Ψ , Aij Ψ1 = ∂ξξj ∂ξi i,j=1

s 1 is equivalent to the H 1 -norm on the subspace of vector functions of Hper (Q) ξ1 orthogonal to the columns of the matrix Φ and this norm corresponds to the inner product s ∂Ψ ∂Θ . , [Ψ, Θ]1 = Aij ∂ξξj ∂ξi i,j=1

Then we see that the functional G(Ψ) is bounded in the norm Ψ1 on this subspace as well, and therefore it can be represented in the form of an inner of theabove subspace. Hence, product [U, Ψ]1 = G(Ψ), where U is an element s

1 on the subspace of vector functions Ψ ∈ Hper such that Φ∗ Ψ = 0, ξ1 (Q) the mentioned above integral identity holds for some vector function U from this subspace. s 1 Let Ψ(ξ) be any vector valued function of space Hper ξ1 (Q) . Represent Ψ(ξ) in the form Ψ(ξ) = Ψ0 (ξ) + Φh, where h is a constant d-dimensional vector and Φ∗ Ψ0 = 0 (h = −Φ∗ ΦΦ∗ Ψ = −Φ∗ Ψ).

Then [U, Ψ]1 = [U, Ψ0 ]1 + [U, Φh]1 = G(Ψ0 ) +

s i,j=1

Aij

∂U ∂ (Φh), ∂ξi ∂ξξj

=

63

2.3. STEADY STATE ELASTICITY EQUATION IN A ROD = G(Ψ0 ) = G(Ψ). Indeed, (F F0 , Φh) −

s

Fj ,

j=1

∂(Φh) ∂ξξj

because Fjk = Fkj ,

= h∗ Φ∗ F0 −

s

Fjk

j,k=1

∂(Φh)k ∂ξξj

=0

∂(Φh)j ∂(Φh)k =− . ∂ξξj ∂ξk

This proves the lemma. Remark 2.3.1. A solution U that is orthogonal to the rigid-body displacements is unique. This follows from the Riesz theorem for G(Ψ). (N,M ) Lemma 2.3.1 implies formula (2.3.14) for the calculation of hl . We introduce now the reflection operator Sα Rs → Rs . For s = 3 we have S2 ξ = (ξ1 , −ξ2 , ξ3 ), S3 ξ = (ξ1 , ξ2 , −ξ3 ). For s = 2 we have S2 ξ = (ξ1 , −ξ2 ). It is assumed that ∀α = 2, . . . , s Sα ((0, 1) × β) = (0, 1) × β. Consider the following conditions: Condition A: ∀α = 2, . . . , s

δαi +δαk +δαj +δαl kl akl aij (ξ). ij (Sα ξ) = (−1)

Figure 2.3.2. Symmetry of a cross-section Condition B: δ1i +δ1k +δ1j +δ1l kl akl aij (ξ), ij (ξ) = (−1)

64


and akl ij is independent of ξ1 . If condition A is satisfied, then the following assertion is valid. Theorem 2.3.1 The matrix hN l is diagonal. If s = 3, then ⎛ ⎛ ⎞ ⎞ ¯ (3) 0 0 E C1 0 0 0 0 ⎜ ⎜ 0 0 0 0⎟ 0 0 0 0 ⎟ ⎜ ⎟ , hN ⎟ hN =⎜ 2 =⎝ 3 ⎝ 0 0 0 0 ⎠, ⎠ 0 0 0 0 ¯ (3) 0 0 0 C2 0 0 0 M ⎛ ⎞ C3 0 0 0 (3) ⎜ 0 −J¯ 0 0⎟ 2 ⎜ ⎟, hN 4 =⎝ (3) ¯ 0 0 −J 0⎠ 3

0

but if s = 2, then ¯ (2) E N h2 = 0

0 , 0

hN 3

=

0

0

C5 0

0 , 0

C4

hN 4

=

C6 0

0 (2) , −J¯2

¯ (s) , J¯(s) , J¯(s) , M ¯ (s) > 0, C1 , C2 , C3 , C4 , C5 , C6 are some constants. where E 2 3 (This theorem as well as following theorems 2.3.2-2.3.7 will be proved in subsection 2.3.6.) Then (2.3.7) takes the form P u(∞) − ΦF ψ =

∞ ∞ d2 ω k 4 3 dl ψ k−2 N d ω 2 Nd ω Nd ω ¯ ψ+ = + µl hM − F µ h +µ h +µh = Φ hN l k 4 3 2 4 3 2 k dx1 dx1 dx1 dxl1 dx1 l=1

k=5

= 0.

(2.3.17)

If condition B is satisfied, then the first sum contains only even powers of µ. Problem (2.3.17) can be regarded as a homogenized equation of infinite order for the d-vector ω. The f.a.s. to this problem is sought as the series ω=

∞

µj ωj (x1 ),

(2.3.18)

j=−2

where ωj (x1 ) are independent of µ, ωj are d-vectors, ωj1 = 0 (ωj4 = 0 if s = 3) for j = −2, −1. The substitution of the series (2.3.18) into (2.3.17) gives the following recurrent chain of equations for the components ωjk of the vectors ωj : ¯ (s) E

d2 ωj1 = fj1 (x1 ), dx21

ωj2 = fj2 (x1 ) dx41

(s) d

−J¯2

4

(s)

and, in addition, for s = 3, −J¯3 d4 ωj3 /dx41 = fj3 (x1 ) ¯ (s) M

d2 ωj4 = fj4 (x1 ), dx21

(2.3.19)

where fjr depend on ωjr1 , j1 < j, and their derivatives. The f.a.s. to problem (2.3.1), (2.3.2), (2.3.4) is thus constructed.

65


2.3.3

Boundary layer corrector.

We construct the f.a.s. to problem (2.3.1)–(2.3.4) in the form, u(∞) = uB + u0P + u1P ,

(2.3.20)


∞

µl Nl0 (ξ)

l=0

u1P =

∞

µl Nl1 (ξ)

l=0

dl ω l+2 0 dl ψ

, + µ Ml (ξ) l dx1 ξ=x/µ dxl1 l=0 ∞

dl ω l+2 1 dl ψ + µ Ml (ξ) l |ξ1 =(x1 −b)/µξ′ =x′ /µ . dxl1 l=0 dx1 ∞

(2.3.21)

Substituting (2.3.20) into (2.3.1)–(2.3.4) and taking into account the fact that uB is as constructed above, we obtain the asymptotic equations ∞

l=0

∞

∞ l l l ¯M x d ψ ¯ lN x d ω + − ΦF¯ ψ − ΦFψ = 0, H µ µl−2 H l µ dxl1 µ dxl1 l=0

¯N µl−1 G l

l=0

x dl ω

µ dxl1 ∞

+

∞

¯M µl+1 G l

l=0

l

¯l ] d ω + µl [N dxl1 l=0

∞ l=0

= 0,

µ dxl1

∞

l

¯ l ] d ψ = 0, µl+2 [M dxl1 l=0

l=0

¯l (0, ξ ′ ) µl N

l=0

∞ l=0

x ∈ Σ, µ

x ∈ Σ, µ

∞ l dl ω

¯ l (0, ξ ′ ) d ψ

= 0, + µl+2 M

l dxl1 x1 =0 dx1 x1 =0 l=0

∞ b dl ψ l

l+2 ¯ ¯l b , ξ ′ d ω

= 0, , ξ′ + µ M µl N

l µ µ dxl1 x1 =b dxl1 x1 =b l=0

¯ N x/µ = H N x/µ + H N 0 x/µ + H N 1 x1 − b/µ, x′ /µ , H l l l l ¯ N x/µ = GN x/µ + GN 0 x/µ + GN 1 x1 − b/µ, x′ /µ , G l l l l ¯l x/µ = Nl x/µ + Nl0 x/µ + Nl1 x1 − b/µ, x′ /µ , N HlN r (ξ) = Lξξ Nlr + TlN r (ξ),

TlN r (ξ) =

s j=1

∂/∂ξξj (Aj1 Nlr−1 ) +

s j=1

(2.3.22)

x ∈ ∂C Cµ ∩ ∂U Uµ , (2.3.23)

∞ x dl ψ l ¯M ¯N x d ω + = 0, µl+1 G µl−1 G l l l µ dxl1 µ dx1

∞

where

x dl ψ

x ∈ Cµ ,

A1j ∂N Nlr−1 /∂ξξj + A11 Nlr−2 ;

(2.3.24)

(2.3.25)

(2.3.26)

(2.3.27)

66

CHAPTER 2. HETEROGENEOUS ROD r GN = l

s

m=1

Amj ∂N Nlr /∂ξξj + Am1 Nlr−1 nm

(2.3.28)

¯M, G ¯M , M ¯ l , T M r , and GM r are obtained by replacing N with M . Here and H l l l l (N,M ) (N,M ) Nl (ξ), Ml (ξ), Hl (ξ), and Gl (ξ) are the 1-periodic functions in ξ1 con(N,M )r (N,M )r r structed above and Nl (ξ), Hl (ξ), and Gl (ξ) exponentially stabilize to zero as ξ1 → ±∞ (+ for r = 0 and − for r = 1). The superscript (N, M ) means that the assertion is true for both N and M . The exponential stabilization to zero implies that r |N Nl (ξ)|dξ, |M Mlr (ξ)|dξ ≤ c1 e−c2 |σ| as σ → ±∞ (σ,σ +1)×β

(σ,σ +1)×β

(as above, + for r = 0 and − for r = 1). Here c1 and c2 are positive constants independent of σ. We arrange Nlr and Mlr in ascending order with respect to the index l and (N,M )r note that Tl is defined by the functions Nlr1 (or Mlr1 ) for l1 < l. We require (N,M )r ¯ ¯ l (0, ξ ′ ), N ¯l (b/ε, ξ ′ ), and M ¯ l (b/ε, ξ ′ ) satisfy , Nl (0, ξ ′ ), M that H l

(N,M )r

Hl

= 0,

(N,M )r

Gl

= 0,

¯ l (0, ξ ′ ) = Φ(0, ξ ′ )hM 0 + O(e−c/µ ), M l

0 ¯l (0, ξ ′ ) = Φ(0, ξ ′ )hN N + O(e−c/µ ), l

1 ξ ′ )hN ¯l b , ξ ′ = Φ(0, + O(e−c/µ ), N l µ

ξ ′ )hM 1 + O(e−c/µ ), ¯ l b , ξ ′ = Φ(0, M l µ

c > 0,

(2.3.29)

where Φ(ξ) is the extended s × d¯ matrix of rigid displacements, 1 0 −ξ2 Φ(ξ) = , 0 1 ξ1 for s = 2 and d¯ = 3, and

⎛

1 0 0 0 Φ(ξ) = ⎝ 0 1 0 −aξ3 0 0 1 aξ2

−ξ2 ξ1 0

⎞ −ξ3 0 ⎠, ξ1

(N,M )r are to be defined from for s = 3 and d¯ = 6. The constant d¯× d matrices hl the conditions of the existence of matrix functions Nlr and Mlr exponentially stabilizing to zero. We construct the functions Nlr , Mlr satisfying equations (2.3.29) by induction on l. Assume that Nlr , Mlr = 0 for l < 0. Let all Nlr1 , Mlr1 be constructed for 0 ¯ hN such that there exists a solution, l1 < l. We choose a constant (d×d)-matrix l exponentially stabilizing to zero, to the problem

Lξξ Nl0 + TlN 0 = 0,

ξ1 > 0,

ξ ′ ∈ β,

2.3. STEADY STATE ELASTICITY EQUATION IN A ROD s ∂N Nl0 =− Am1 Nl−1 nm , ∂ννξ m=2

67

ξ ′ ∈ ∂β,

0 ξ ′ )hN Nl0 (0, ξ ′ ) = −N Nl (0, ξ ′ ) + Φ(0, l ,

(2.3.30)

1 hN l

and choose a constant matrix such that there exists a solution exponentially stabilizing to zero to the problem Lξξ Nl1 + TlN 1 = 0,

ξ1 < 0,


ξ ′ ∈ β,

ξ ′ ∈ ∂β,

1 ξ ′ )hN Nl1 (0, ξ ′ ) = −N Nl (0, ξ ′ ) + Φ(0, l .

Mlr

(2.3.31)

r hM l

The matrices and are defined similarly (by replacing N with M in equations (2.3.30) and (2.3.31)). The following theorem provides the possibility of such a choice of the con(N,M )r . stants hl Theorem 2.3.2. Let Aij (ξ) satisfy the above conditions, let s-vectors Fj (ξ) ∈ L2 ([0, +∞) × β), j = 0, 1, . . . , s, satisfy |F Fj (ξ)|dξ ≤ c1 e−c2 σ , σ > 0, (2.3.32) (σ,σ +1)×β

and let an s-vector u0 (ξ ′ ) ∈ H 1/2 ({0}×β) be given. Then there exists a solution u(ξ) to the problem Lξξ u = F0 (ξ) +

s ∂F Fj j=1

∂ξξj

,

ξ1 > 0,

ξ ′ ∈ β,

(2.3.33)

s

∂u = Fj nj , ξ ′ ∈ ∂β, ∂ννξ j=2

u ξ1 =0 = u0 (ξ ′ )

(2.3.34) (2.3.35)

such that for some rigid displacement w(ξ) = Φ(ξ)h, where h is a constant ¯ d-vector, the following inequality holds |u − w|dξ ≤ c1 e−c2 σ , c1 , c2 > 0, σ > 0. (2.3.36) (σ,σ +1)×β

Remark 2.3.2. Let us explain some terms from Theorem 2.3.2 (see [129],[130]). The space H 1/2 ({0} × β) is thought of as the space of traces V (ξ ′ ) on the set {ξ1 = 0, ξ ′ ∈ β} of vector functions from H 1 ((0, 1)×β) equipped with the norm

V 21/2 = inf e(v) + |v|2 dξ, v ∈ W21 ((0, 1) × β), v ξ1 =0 = V (ξ ′ ) , v

(0,1)×β

68


e(v) =

s ∂v j 2 ∂v i , + ∂xi ∂xj i,j=1

v = (v 1 , . . . , v s )∗ .

A solution to problem (2.3.33)–(2.3.35) is viewed as an s-vector function u(ξ) ∈ H 1 loc whose trace is equal to u0 (ξ ′ ) for ξ1 = 0. For any vector function . Ψ(ξ) ∈→ C ∞ with support in the cylinder (δ, D) × β, δ > 0, the function u(ξ) must satisfy the integral identity s ∂u ∂Ψ , dξ = (F F0 , Ψ) dξ− − Aij ∂ξξj ∂ξi (0,+∞)×β i,j=1 (0,+∞)×β

−

s j=1

(0,+∞)×β

Fj ,

∂Ψ dξ. ∂ξξj

Theorem 2.3.2 follows directly from Theorems 4 and 5 in [130]. Their proofs are “corrected” for the right-hand side having specific representation (2.3.33) and (2.3.34) and the momenta in Theorem 5 are chosen so that they are equal to zero at infinity (i.e., in the notation of [130], the momenta are s ∂η r P r (0, u) = − (F F0 , η r )dξ + Fj , dξ, ∂ξξj (0,+∞)×β (0,+∞)×β j=1 where η r , r = 1, . . . , d¯ are elements of the basis of the rigid displacements space). 0 To determine the matrix hN from (2.3.30), it suffices to first solve the l problem 0 + T N 0 = 0, ξ1 > 0, ξ ′ ∈ β, Lξξ N l l s 0 ∂N l =− nm Am1 Nl0−1 , ∂ννξ m=2

ξ ′ ∈ ∂β,

l0 (0, ξ ′ ) = −N N Nl (0, ξ ′ ).

(2.3.37)

0 stabilizes to According to Theorem 2.3.2, each vector-column of the matrix N l 0 a rigid displacement, so that the entire matrix Nl (ξ) stabilizes to some matrix where h is a constant d¯ × d-matrix. If one subtracts Φh from N 0 (ξ), then Φh, l 0 (ξ) − Φh will stabilize to zero and satisfy (2.3.30) for the difference Nl0 = N l 0 r hN = −h. Problem (2.3.31) and the problems for Mlr and hM can be solved l l in the same manner. It follows from [130] that the energy e of the columns Nlr and Mlr also exponentially stabilizes to zero. After the successive determination of all Nlr and Mlr by induction (note that r N0 = 0) we obtain a homogenized problem of infinite order for the d-vector function ω(x1 ). This problem is given by equation (2.3.17) with the boundary conditions ∞ ∞ l

l M r(cut d) d ψ N r(cut d) d ω = 0, (2.3.38) + µl+2 hl ω+ µl hl

l dxl1 x1 =rb dx1 l=0 l=1

69

2.3. STEADY STATE ELASTICITY EQUATION IN A ROD ∞ ∞ d(ω 2 , . . . , ω s )∗ l

l N r(oth) d ω l+2 M r(oth) d ψ = + µ h + µl hl µ

l dx1 dxl1 x1 =rb dxl1 l=0

l=2

= 0,

r = 0, 1,

(2.3.39)

where the superscript (cut d) denotes the first d rows of a matrix and the superscript (oth) denotes the remaining d¯ − d rows (we recall that d = 2 and d¯ = 3 if s = 2; d = 4 and d¯ = 6 if s = 3); ω 1 , ω 2 , . . . , ω d are the components of the vector ω. r r It is significant that the matrices hN and hN have the following structure: 0 1 for s = 2: ⎞ ⎛ 1 0 r ⎝ 0 1⎠ , hN = 0 0 0 r is equal to (0, 1). and the third row of hN 1 ⎛ 1 0 ⎜ 0 1 ⎜ ⎜ 0 0 r ⎜ hN = 0 ⎜ 0 0 ⎜ ⎝ 0 0 0 0

For s = 3 we have ⎞ 0 0 0 0⎟ ⎟ 1 0⎟ ⎟, 0 1⎟ ⎟ 0 0⎠ 0 0

r and the fifth and sixth rows of hN are equal to (0, 1, 0, 0) and (0, 0, 1, 0), re1 spectively. The proof of this assertion (Lemma 2.3.6) will be given in the sequel. The f.a.s. to problem (2.3.17), (2.3.38) and (2.3.39) is represented in the form of the series (2.3.18) and, being substituted into (2.3.17), (2.3.38) and (2.3.39), leads to the recurrent chain of equations (2.3.19) for the components ωlk of the vectors ωj with the boundary conditions

ωj x =rb = gjr , r = 0, 1, 1

dωjq

d+q = gjr ,

dx1 x1 =rb

r = 0, 1,

q = 2, . . . , s.

(2.3.40)

d+q , q = 1, . . . , d¯ depend on ωjr1 when j1 < j, and on the derivatives Here fjr , gjr of these functions. Thus, the f.a.s. to problem (2.3.1)–(2.3.4) is constructed.

2.3.4

The boundary layer corrector when the left end of the bar is free

Consider problem (2.3.1)–(2.3.3) with condition (2.3.3) if x1 = 0, replaced by the free surface condition s x ∂u ∂u = 0. ≡− A1j µ ∂xj ∂ν j=1

(2.3.3′ )

70


Then, when constructing the boundary layer corrector, condition (2.3.26) must be replaced by ∞ l=0

∞ ′ l ′ l ¯ M 0, x d ψ

¯ N 0, x d ω

=0 + µl+1 G µl−1 G l l l µ dxl1 x1 =0 µ dx1 x1 =0

(2.3.26′ )

l=0

¯l (0, ξ ′ ) and M ¯ l (0, ξ ′ ) must be replaced by and the requirements (2.3.29) on N ′ ′ ¯N N 0 + O(e−c/µ ), G ¯M M 0 + O(e−c/µ ), G l (0, ξ ) = Φhl l (0, ξ ) = Φhl

c > 0. (2.3.29′ ) r r The construction of the functions Nl and Ml is carried out, as above, by induction on l; however, problem (2.3.30) is replaced by Lξξ Nl0 + TlN 0 = 0,

ξ ′ ∈ β,

ξ1 > 0,


ξ ′ ∈ ∂β,

∂N Nl0 ∂N Nl 0 ξ ′ )hN + A11 (N Nl0−1 + Nl−1 ) + Φ(0, =− l , ∂ννξ ∂ννξ

ξ1 = 0,

0 must be chosen so as to ensure that there where the d¯ × d constant matrix hN l 0 exists a solution Nl exponentially stabilizing to zero. The matrices Ml0 and 0 hM are defined in the usual way. l (N,M )0 is a consequence of the The possibility of such a choice of constants hl following theorem. Theorem 2.3.3. Let Aij (ξ) satisfy the conditions imposed at the beginning of the section, let the s-vectors

Fj (ξ) ∈ L2 ([0, +∞) × β),

j = 0, 1, . . . , s,

¯ satisfy (2.3.32), let an s-vector u0 (ξ ′ ) ∈ L2 (β), and let the d-dimensional constant vector h be given by 1 ∗ ′ ′ −1 ∗ (ξ)F Φ F0 (ξ) dξ− h = Φ (0, ξ )Φ(0, ξ ) mes β (0,+∞)×β

∗ (0, ξ ′ ) × u0 (ξ ′ ) . −Φ

(2.3.41)

Then there exists a solution u(ξ) to the problem Lξξ u = F0 (ξ) +

s ∂F Fj j=1

∂ξξj

,

ξ1 > 0,

ξ ′ ∈ β,

(2.3.42)

s

∂u = Fj nj , ∂ννξ j=2

ξ ′ ∈ ∂β,

(2.3.43)

2.3. STEADY STATE ELASTICITY EQUATION IN A ROD ∂u ξ ′ )h, ξ1 = 0 = −F F1 + u0 (ξ ′ ) + Φ(0, ∂ννξ exponentially stabilizing to zero, |u| dξ ≤ c1 e−c2 σ , c1 , c2 > 0.

71

(2.3.44)

(σ,σ +1)×β

The solution to problem (2.3.42)–(2.3.44) is thought of as an s-vector function u(ξ) ∈ H 1 loc satisfying the following integral identity for any vector func¯ equal to zero for sufficiently large ξ1 : tion Ψ(ξ) ∈ C ∞ ([0, +∞) × β) s ∂u ∂Ψ dξ = , − Aij ∂ξξj ∂ξi (0,+∞)×β i,j=1

=

(0,+∞)×β

(F F0 , Ψ) dξ − −

β

s j=1

(0,+∞)×β

Fj ,

∂Ψ dξ− ∂ξξj

ξ ′ )h + u0 , Ψ) dξ ′ . (Φ(0,

Theorem 2.3.3 follows from Theorem 4 and from an analog of Theorem 5 in [130]. It follows from Theorem 2.3.3 that s 1 ∂N Nl0−1 N0 ∗ ′ ′ −1 ∗ (ξ) Φ A1j + = −Φ (0, ξ )Φ(0, ξ ) hl mes β (0,+∞)×β ∂ξξj j=1 s ∂N Nl ∗ (0, ξ ′ ) × + A11 Nl−1 + A11 Nl0−2 dξ + Φ A1j ∂ξξj

(2.3.45)

j=1

0 and a similar equation holds for hM l . Thus, (2.3.26) takes the form

Φ

∞ l=0

0 µl−1 hN l

∞ l dl ω

M0 d ψ = 0. + µ h

l+1 l dxl1 x1 =0 dxl1 x1 =0 l=0

(2.3.46)

We have N00 = 0, T1N 0 = 0. The second (and the third if s = 3) row of the matrix N10 is a solution to the homogeneous problem and therefore is equal to zero. Hence, the second (and the third) row of T2N 0 is zero as well. In what follows conditions A are assumed to be satisfied. N 0(cut d) 0 Theorem 2.3.4. The matrices hl are diagonal, and hN = 0. For 0 s = 3 we have ⎞ ⎛ ¯ (3) 0 0 E 0 ⎜ 0 0 0 0 ⎟ ⎜ ⎟ ⎜ 0 0 0 0 ⎟ N0 ⎜ ⎟ h1 = − ⎜ ¯ (3) ⎟ , 0 0 0 M ⎜ ⎟ ⎝ 0 0 0 0 ⎠ 0 0 0 0

72


0 hN 2

⎛

⎜ ⎜ ⎜ = −⎜ ⎜ ⎜ ⎝

N 0(cut d)

h3

⎞ 0 0 0 0 0 0⎟ ⎟ 0 0 0⎟ ⎟ 0 0 C2 ⎟ , ⎟ (3) J¯2 /ξ22 0 0⎠ (3) 0 J¯3 /ξ32 0

C1 0 0 0 0 0 ⎛

C3 ⎜ 0 = −⎜ ⎝ 0 0

0 (3) −J¯

0 0 (3) −J¯3 0

2

0 0

⎞ 0 0⎟ ⎟, 0⎠ C4

¯ (3) , M ¯ (3) , J¯(3) , J¯(3) > 0 are the constants from Theorem 1, and C1 , where E 2 3 C2 , C3 , C4 are some constants. For s = 2 we have ⎛ ⎞ ¯ (2) 0 E 0 ⎝ hN 0 0⎠ , 1 =− 0 0 0 hN 2

⎛

C5 = −⎝ 0 0

N 0(cut d) h3

=−

⎞ 0 ⎠, 0 (2) J¯2 /ξ22

C6 0

0 (2) , −J¯ 2

(2)

¯ (2) , J¯ > 0 are the constants from Theorem 1, and C5 , C6 are some where E 2 constants. It follows from Theorem 2.3.4 that upon substituting the series (2.3.18) into the homogenized problem of infinite order {(2.3.17), (2.3.38) for r = 1 (i.e., x1 = b) and (2.3.46)} we obtain a recurrent chain of problems for the functions ωj . They consist of equation (2.3.19), for x1 = b of the boundary conditions (2.3.40) (r = 1), and for x1 = 0 of the boundary conditions ¯ (s) E

dωj1 1 = gj0 , dx1

ωj2 2 = gj0 , dx31

(s) d

−J¯2

3

(s) J¯2 d2 ωj2 3 = gj0 ξ22 dx21

and, moreover, of ωj3 4 = gj0 , dx31

(s) d

−J¯3

3

¯ (s) M

dωj4 5 = gj0 , dx1

k if s = 3, where gj0 depend on ωj1 if j1 < j.

(s) J¯3 d2 ωjs 6 = gj0 ξ32 dx21

(2.3.47)


2.3.5

73

The boundary layer corrector for the two bar contact problem

Consider the following problem, which is, in some sense, a model problem for systems consisting of a finite number of bars. Namely, let the two bars Cµ+ = Uµ ∩ {x1 ∈ (0, b1 )} and Cµ− = Uµ ∩ {x1 ∈ (−b2 , 0)}, be given, where b1 and b2 are integer multiples of µ and each bar has specific coefficients Aij (ξ), i.e., Aij (ξ) = A+ ij (ξ)

if ξ1 > 0,

Aij (ξ) = A− ij (ξ)

if ξ1 < 0,

where A± ij (ξ) are 1-periodic in ξ1 matrix functions satisfying the conditions stated at the beginning of this section. Consider the elasticity system of equaUµ and tions (2.3.1) on Cµ+ ∪ Cµ− with the boundary conditions (2.3.2) for x ∈ ∂U (2.3.3) for x1 = b1 or −b2 , i.e., P u = Fµ (x),

x ∈ Cµ+ ∪ Cµ− ,

∂u Uµ , = 0, x ∈ ∂U ∂ν u = 0 when x1 = b1 or x1 = −b2 .

(2.3.48)

Figure 2.3.3 Two rod system On the discontinuity interfaces of the coefficients (in particular, for x1 = 0) the conjugation conditions (2.3.4) are imposed. Such a problem models the stress-strain state of a bar consisting of two parts with different microstructure.

74


The construction of a f.a.s. to problem (2.3.48), (2.3.4) is again carried out in two stages. In the first stage, we construct inner expansions for Cµ+ and Cµ− , and in the second stage we construct the boundary layer correctors. As above, they are constructed independently for the cross-sections {x1 = −b2 }, {x1 = b1 }, and {x1 = 0}. The construction of the boundary layers near the ends {x1 = −b2 } and {x1 = b1 } is described above (see item 2.3). In what follows, we dwell on the construction of the boundary layer corrector near the cross-section {x1 = 0}. We seek the f.a.s. in the form −1 +1 0 u(∞) = u− B + uP + u P + u P

for x1 < 0,

−1 +1 0 u(∞) = u+ B + uP + u P + u P

for x1 > 0,

(2.3.49)

u− B

where and u+ B the functions u−1 P

are the corresponding inner decompositions of type (2.3.6), and u+1 P are the boundary layer correctors near the ends {x1 = −b2 } and {x1 = b1 }, respectively (see (2.3.21)), and u0P is the boundary layer corrector near the contact interface u0P =

+

dl ω

dl ω

+ Nl−0 (ξ) l µl Nl+0 (ξ) l dx1 x1 =−0 dx1 x1 =+0 l=0

∞

dl ψ

dl ψ

+ Ml−0 (ξ) l . µl+2 Ml+0 (ξ) l

x =+0 x =−0 ξ=x/µ dx dx 1 1 1 1 l=0

∞

(2.3.50)

The substitution of (2.3.49), (2.3.50) into (2.3.48), (2.3.4) gives the asymptotic equations ∞ l=0

+

l l l N −0 x d ω ¯ lN x d ω + H N +0 x d ω

+ H µl+2 ( H

l l µ dxl1 x1 =−0 µ dxl1 x1 =+0 µ dxl1

∞ l=0

l l l M −0 x d ψ ¯ lM x d ψ + H M +0 x d ψ

+ H µl ( H

l l µ dxl1 x1 =−0 µ dxl1 x1 =+0 µ dxl1 −ΦF¯ ψ − ΦFψ = 0,

x ∈ Cµ+ ∪ Cµ− ;

(2.3.51)

G = l l x dl ω −0 x d ω +0 x d ω ¯N + GN + GN = µl−1 ( G

l l l l l l µ dx1 x1 =−0 µ dx1 x1 =+0 µ dx1 l=0 ∞

+

∞ l=0

l l ¯ M x d ψ + GM +0 x d ψ

+ µl+1 ( G l l l µ dxl1 x1 =+0 µ dx1 −0 +GM l

x dl ψ

,

l µ dx1 x1 =−0

(2.3.52)

75

2.3. STEADY STATE ELASTICITY EQUATION IN A ROD ∞

µl

l=0

+

∞

µl+2

l=0

l l l −0 d ω +0 d ω ¯l d ω + [N + [N N ] N ] N

l l dxl1 x1 =−0 dxl1 x1 =+0 dxl1

l l l −0 d ψ +0 d ψ ¯ l d ψ + [M + [M M ] = 0, (2.3.53) M M ]

l l dxl1 x1 =+0 dxl1 x1 =−0 dxl1

x ∈ Σ or x1 = 0, µ ∞ dl ω l l −0 d ω +0 d ω ¯l + N + N µl ( N

l l dxl1 x1 =−0 dxl1 x1 =+0 dxl1 l=0 [G] = 0,

+

dl ψ dl ψ

dl ψ

¯l = 0, + Ml−0 l + Ml+0 l µl+2 ( M l dx1 x1 =−0 dx1 x1 =+0 dx1 l=0

∞

(2.3.54)

(2.3.55)

for ξ1 = −b2 /µ, x1 = −b2 and for ξ1 = b1 /µ, x1 = b1 , where

′ ′ ¯l x = Nl x + N − x1 + b2 , x + N + x1 − b1 , x , N l l µ µ µ µ µ µ

Nl (ξ)

are the matrix valued functions of the inner expansion (specific for each of the subspaces ξ1 > 0 and ξ1 < 0), while Nl− (ξ), Nl+ (ξ) are the boundary layer correctors near the ends of the bar satisfying estimates of type (2.3.28). The ¯ (N,M ) and G ¯ (N,M ) have a similar structure. The functions N ±0 (ξ), matrices H l l l (N,M )±0 (N,M )±0 ±0 Ml (ξ), Hl , and Gl exponentially stabilize to zero as |ξ1 | → +∞ (in the sense of (2.3.28)). Here HlN ±0 = Lξξ Nl±0 ,

±0 GN (ξ) = l

s

Amj

m,j=1

∂N Nl±0 nm , ∂ξξj

¯M, G ¯ M , Ml , H M ±0 , GM ±0 , M ±0 , GM ±0 are obtained by replacing N and H l l l l l l with M . We require that the conditions (N,M )±0

Hl

(N,M )±0

(ξ) = 0,

(N,M )±

Hl

(N,M )±

(ξ) = 0,

(ξ) = 0, ξ ′ ∈ ∂β, ξ ′ )hN − +O e−c/µ , M ξ ′ )hM − +O e−c/µ , ¯l − b2 , ξ ′ = Φ(0, ¯ l − b2 , ξ ′ = Φ(0, N l l µ µ ξ ′ )hM + +O e−c/µ , ξ ′ )hN + +O e−c/µ , M ¯ l b1 , ξ ′ = Φ(0, ¯l b1 , ξ ′ = Φ(0, N l l µ µ

+0 ′ Nl (+0, ξ ) + [N Nl ] ξ =0 = 0, −N Nl (−0, ξ ′ ) + [N Nl−0 ] ξ =0 = 0, 1 1

+0 −0 ′ ′ Ml (+0, ξ ) + [M Ml ] ξ1 =0 = 0, −M Ml (−0, ξ ) + [M Ml ] ξ1 =0 = 0,

(N,M ) (N,M )+0 Gl (+0, ξ ′ ) + [Gl ] ξ1 =0 = 0, Gl

(ξ) = 0,

Gl

76

CHAPTER 2. HETEROGENEOUS ROD (N,M )

−Gl

(N,M )−0

(−0, ξ ′ ) + [Gl

and the conditions

] ξ1 =0 = 0

[N Nl±0 ] = 0,

[N Nl± ] = 0,

ξ ∈ Σ,

[G±0 l ] = 0,

[G± l ] = 0,

ξ∈Σ

(2.3.56)

(2.3.57)

hold. For Nl and Ml we obtain analogs of problems (2.3.11)-(2.3.14) (specific for ξ1 < 0 and ξ1 > 0) and analogs of problems (2.3.30), (2.3.31) for the functions Nl± , Ml± . Finally, for Nl± we obtain the following problems in the cylinder R × β: Lξξ Nl±0 = 0, ξ ∈ R × β,

[N Nl±0 ] ξ

∂ N ±0 = 0, ∂ννξ l

1 =0

ξ ′ ∈ ∂β,

¯ N ±0 , ξ ′ )h = ∓N Nl (±0, ξ ′ ) + Φ(0, l

∂N ∂N Nl±0

Nl ¯ N ±0 , (2.3.58) ξ ′ )h (±0, ξ ′ )+A11 Nl−1 (±0, ξ ′ )n1 + Φ(0, =∓

l ∂ννξ ∂ννξ ξ1 =0

¯ N ±0 are chosen from the ¯ N ±0 and h where n1 > 0, the d¯ × d constant matrices h l l condition of existence of a solution exponentially stabilizing to zero. Here ¯ N ±0 = h l

N l ∗ (0, ξ ′ ) ± ∂N ∗ (0, ξ ′ )Φ(0, ξ ′ )−1 Φ = Φ (±0, ξ ′ ) + A11 Nl−1 (±0, ξ ′ )n1 . ∂ννξ (2.3.59) Equation (2.3.59) follows from Theorem 2.3.5. Theorem 2.3.5. Let Aij (ξ) satisfy the conditions stated at the beginning of the section for ξ1 < 0 and ξ1 > 0, let the s-dimensional vectors Fj (ξ) ∈ L2 (IIR × β), j = 0, . . . , s satisfy estimates (2.3.32), let s-dimensional vectors u0 (ξ ′ ) ∈ H 1/2 ({0} × β), u1 (ξ ′ ) ∈ L2 (β), and let the d-dimensional constant vector be given by 1 ¯ = −Φ ∗ (ξ)F ∗ (0, ξ ′ )u1 (ξ ′ ) . ξ ′ )−1 ∗ (0, ξ ′ )Φ(0, Φ F0 (ξ) dξ + Φ h mes β IR×β (2.3.60) Then there exists a solution u(ξ) to the problem

Lξξ u = F0 (ξ) +

s ∂F Fj j=1

∂ξξj

,

ξ1 > 0,

ξ ′ ∈ β,

(2.3.61)

s

∂u = Fj nj , ∂ννξ j=2 [u] = u0 (ξ ′ ),

ξ ′ ∈ ∂β, ξ1 = 0,

(2.3.62) (2.3.63)

77

2.3. STEADY STATE ELASTICITY EQUATION IN A ROD s

A1j

j=1

∂u ¯ ξ ′ )h, = [F F1 ] + u1 (ξ ′ ) + Φ(0, ∂ξξj

ξ1 = 0

(2.3.64)

¯ where h ¯ is a constant h, such that for some rigid displacement w(ξ) = Φ(ξ) ¯ d-dimensional vector, inequality (2.3.36) holds, and, in addition,

(σ,σ +1)×β

|u| dξ ≤ c1 e−c2 σ ,

σ < −1.

Here the normal to the discontinuity interface ξ1 = 0 is co-directed with the Oξ1 -axis. The solution to problem (2.3.61)–(2.3.64) is thought of as an svector function u(ξ) ∈ H 1 loc satisfying the following integral identity for any ¯ equal to zero for ξ1 sufficiently large: vector-function Ψ(ξ) ∈ C ∞ (IIR × β) −

s

i,j=1

IR×β

∂u ∂Ψ dξ = , Aij ∂ξξj ∂ξi

+

β

IR×β

(F F0 , Ψ) dξ −

s j=1

IR×β

Fj ,

∂Ψ dξ ∂ξ

¯ + u , Ψ) dξ ′ . ξ ′ )h (Φ(0, 1

¯ is carried out as that of the constant hN 0 in problem The choice of the constant h (2.3.30). By analogy with the problems (2.3.1)–(2.3.4) and (2.3.48), from (2.3.51)– (2.3.55) we obtain the homogenized problem: equations (2.3.17) with coeffiM cients hN l and hl constant on each half-line R− = (−∞, 0) and R+ = (0, +∞), boundary conditions (2.3.38), (2.3.39) at x1 = −b2 and x1 = b1 , and the interface conditions at x1 = 0, [ω] +

∞

ω

+

dxl1 x1 =±0

N ±)(cut d) d

¯ µl h l

l=1 ±

+

∞

¯ M ±0(cut d) µl+2 h

l=0 ±

l

dl ψ

= 0,

dxl1 x1 =±0

(2.3.65)

∞ d l ¯ N ±O(oth) d ω

(ω 2 , . . . , ω s )∗ + µl h µ l dx dxl1 x1 =±0 ± l=2

+

∞ l=0

∞ l=0 ±

µl−1 ¯¯hl N ±0

ψ

= 0,

dxl1 x1 =±0

M ±O(oth) d

¯ µl+2 h l

l

∞ dl ψ

dl ω

= 0. + µl+1 ¯¯hl M ±0 l

l dx1 x1 =±0 dx1 x1 =±0 l=0 ±

(2.3.66)

(2.3.67)

78


¯ N ±0 = 0. Theorem 2.3.6. The matrices ¯¯hl N ±O(cut d) are diagonal, and h 0 For s = 3 we have ⎛ ⎞ ¯ (3) 0 0 E 0 ± ⎜ 0 0 0 0 ⎟ ⎟ ⎜ ⎜ 0 0 0 0 ⎟ N ±0 ¯ ⎟, = ±⎜ h1 ⎜ ¯ (3) ⎟ 0 0 0 M ⎜ ± ⎟ ⎝ 0 0 0 0 ⎠ 0 0 0 0 ⎛

¯ N ±0 h 2

⎜ ⎜ ⎜ ⎜ = ±⎜ ⎜ ⎜ ⎜ ⎝

¯ N ±0(cut d) h 3 and for s = 2 we have ¯ N ±0 h 1

C1 0 0 0

0 0 0 0

0

(3) J¯2±

ξ22

0

⎛

C3 ⎜ 0 = ±⎜ ⎝ 0 0

0

0 0 0 0

0 (3) J¯2±

ξ32

0 (3) −J¯2± 0 0

⎞ 0 0⎠ ,

⎛

¯ (2) E ± = ±⎝ 0 0

⎞ 0 0⎟ ⎟ 0⎟ ⎟ C2 ⎟ , ⎟ ⎟ 0⎟ ⎠ 0

0 0 (3) −J¯3± 0 ⎛

C5 ¯ N ±0 = ± ⎝ 0 h 2 0 0 C6 0 ¯ N ±0(cut d) = ± h (2) , 3 0 −J¯2±

⎞ 0 0⎟ ⎟ 0⎠ C4 ⎞ 0 0 ⎠, (2) J¯ 2±

¯ (s) , M ¯ (s) , J¯(s) , J¯(s) > 0 are the constants from Theorem 1 (the sign + where E ± ± 2± 3± ¯ N ±0 are calculated or - denotes the half-line R+ or R− , where the coefficients h l ± by Aij (ξ)), C1 , C2 , C3 , C4 , C5 , C6 are certain constants. As before, we seek the f.a.s. of a homogenized problem of infinite order represented as the series (2.3.18) and obtain a recurrent chain of problems for the function ωj : equations (2.3.19) for (−b2 , 0) and (0, b1 (with their specific coefficients on each half-line), boundary conditions (2.3.40) at x1 = −b2 and x1 = b1 and interface conditions at x1 = 0 of the form 1 [ωj ] = gj0 ,

dω q j

dx1

2q = gj0 ,

¯ (s) E

q = 2, . . . , s,

dωj1 3 = gj0 , dx1

79


(s) − J¯2

d3 ωj2 4 = gj0 , dx31

J¯(s) d2 ω 2 j 2

ξ22 dx21

When s = 3 we add also

5 = gj0 ,

ωj3 6 = gj0 , dx31

(s) d

− J¯3

3

4 7 ¯ (s) d ωj = gj0 , M dx1 J¯(s) d2 ω 3 j 8 3 = gj0 . ξ32 dx21

(2.3.68)

¯ (s) , M ¯ (s) , J¯(s) and J¯(s) for R+ at the point x1 = +0 Here we take the values E 2 3 k and for R− at the point x1 = −0, gj0 depend on ωj1 for j1 < j.

2.3.6

Homogenized problem of zero order

The leading term of the homogenized problem of infinite order as µ → 0 will be called a homogenized problem of zero order . From (2.3.17) we obtain ℜµ v = F¯ ψ, (2.3.69) where

⎛

⎜ ⎜ ℜµ = ⎜ ⎝

¯ (3) ∂ 2 E 0 0 0

for s = 3 and for s = 2; ℜµ =

0 (3) −µ2 J¯ ∂ 4 2

0 0

¯ (2) ∂ 2 E 0

0 0 2 ¯(3) 4 −µ J3 ∂ 0

0 , (2) −µ2 J¯2 ∂ 4

0 0 0

¯ (3) ∂ 2 M

∂=

⎞ ⎟ ⎟ ⎟ ⎠

d dx1

v is the unknown vector function in (2.3.69). From (2.3.39) we obtain V = 0,

x1 = 0, b,

where for s = 3 we define V = (v 1 , v 2 , v 3 , v 4 , µ∂v 2 , µ∂v 3 )∗ , and for s = 2 we define V = (v 1 , v 2 , µ∂v 2 )∗ .

(2.3.70)

80


and v q are the components of the vector v. Thus, the homogenized problem of zero order for (2.3.1)–(2.3.4) is problem (2.3.69),(2.3.70). The boundary conditions of second kind give (2.3.46), whence follows −Λµ v = 0, where

⎛

for s = 3 and

⎜ ⎜ ⎜ ⎜ ⎜ Λµ = ⎜ ⎜ ⎜ ⎜ ⎝

¯ (3) ∂ E 0 0 0

0 (3) −µ2 J¯ ∂ 3

0

µ ξ22 ∂ 2

2

0 0

0 0 2 ¯(3) 3 −µ J3 ∂ 0

(3) J¯

0

2

0

(3) J¯ µ ξ32 ∂ 2 3

0

⎛

¯ (2) ∂ E ⎜ 0 Λµ = ⎝ 0

(2.3.71)

⎞ 0 (2) −µ2 J¯2 ∂ 3 ⎟ ⎠ µ

0 0 0

⎞

⎟ ⎟ ⎟ ⎟ ⎟ (3) ⎟ ¯ M ∂⎟ ⎟ 0 ⎟ ⎠ 0

(2) J¯2 ∂ 2

ξ22

for s = 2. Thus, the homogenized problem of zero order for (2.3.1)–(2.3.4), (2.3.3′ ) has the representation (2.3.69), (2.3.70) if x1 = b and (2.3.71) if x1 = 0. The interface conditions at x1 = 0 yield the homogenized interface conditions of zero order

−

[V ] = 0, x1 = 0, Λ+ (2.3.72) µ v x =+0 − Λµ v x =−0 = 0, 1

1

− ¯ (s) , J¯r(s) , M ¯ (s) with E ¯ (s) , J¯(s) , where Λ+ µ and Λµ are obtained by replacing E + r+ − ¯ (s) or E ¯ (s) , J¯(s) , M ¯ (s) , respectively. We define the operators ℜ+ M and ℜ µ µ in + − r− − a similar way. Then problem (2.3.48) gives the homogenized problem of zero order ¯ ℜ+ x1 ∈ (0, b1 ), µ v = F ψ,

¯ ℜ− µ v = F ψ,

x1 ∈ (−b2 , 0),

V = 0 when x1 = b1 , x1 − b2 ,

−

[V ] = 0 when x1 = 0, Λ+ µ v x =+0 − Λµ v x =−0 = 0. 1

1

(2.3.73)

To calculate the coefficients of the operator ℜµ , we must solve the cell problems s ∂ ∂ (1) (N N1 + eξ1 ) = 0, ξ ′ ∈ β, Amj (ξ) ∂ξξj ∂ξm m,j=1

∂ (1) (N N + eξ1 ) = 0, ∂ννξ 1

ξ ′ ∈ ∂β,

(2.3.74)

81


∂ ∂ (4) (N N1 + ηξ1 ) = 0, Amj (ξ) ∂ξξj ∂ξm m,j=1 s

(1)

∂ (4) (N N + ηξ1 ) = 0, ∂ννξ 1

ξ ′ ∈ β,

ξ ′ ∈ ∂β,

(2.3.75)

(4)

where N1 and N1 are 1-periodic in ξ1 s-vector functions with components k4 ∗ ∗ ∗ nk1 1 and n1 , e = (1, 0, 0) for s = 3, e = (1, 0) for s = 2 and η = (0, −ξ3 , ξ2 ) . On solving these equations, we obtain the longitudinal stiffness ¯ (s) = E

s

a1k 1j

j,k=1

∂ (nk1 1 + δk1 ξ1 ) ∂ξξj

(2.3.76)

∂ k (nk4 + η ξ ) . 1 ∂ξξj 1

(2.3.77)

and the torsional stiffness ¯ (s) = M

s

j,k=1

(2)

a1k 1j

(s)

Further, if N2 , . . . , N2 are s-dimensional vector-functions one-periodic in ξ1 ks with components nk2 2 , . . . , n2 , which are solutions to the problem

∂ ∂ (q) (N N2 − Am1 (ξ)eξq ) = 0, Amj (ξ) ∂ξξj ∂ξm m,j=1 s

∂ (q) N2 − Am1 (ξ)eξq ) = 0, nm Amj (ξ) ∂ξξj m,j=1 s

q = 2, . . . , s,

ξ ′ ∈ β,

ξ ′ ∈ ∂β,

(2.3.78)

then the bending stiffness is calculated by means of the formula s ∂ kq n2 − a11 J¯q(s) = − a1k 11 ξq ξq . 1j ∂ξξj

(2.3.79)

j,k=1

Indeed, kq kq qk kq qk ∂n3 1k kq 1k ∂n3 + a n = a + a n −J¯q(s) = hqq = a q1 qj 2 11 2 4 1j ∂ξξj ∂ξξj

=

∂ ∂nkq ∂nkq kq kq ∂ξq 3 3 ξq + a1k a1k = + a1k m1 n2 mj m1 n2 ∂ξξj ∂ξm ∂ξm ∂ξξj ∂n2kq ∂nkq kq 2 + a11 ξq = a1k + a1k = a1k 11 ξq ξq . 11 n1 1j 1j ∂ξξj ∂ξξj

a1k mj

Here nkq l are the components of the matrix Nl . Consider the isotropic case, in which

akl ij (ξ) = (δij δkl + δil δjk )M (ξ) + λ(ξ)δik δjl ,

82


where λ(ξ), M (ξ) are scalar 1-periodic in ξ1 functions (Lam´ é coefficients). If the coefficients λ and M are constant, then the following theorem holds true: Theorem 2.3.7. The following equalities hold: ¯ (3) = M (3λ + 2M ) , E λ+M

¯ (3) M

(2.3.80)

¯ (3) ξ 2 , r = 2, 3, J¯r(3) = E r ∂Y 2 ∂Y 2 , + =M 1− ∂ξ3 ∂ξ2

(2.3.81)

(2.3.82)

where Y (ξ ′ ) is the solution to the Laplace equation ∆Y = 0 in β with the boundary condition ∂Y − ξ3 n2 a + ξ2 n3 a = 0 ∂n

on ∂β, and 2 2 ¯ (2) = (λ + 2M ) − λ , E λ + 2M

(2.3.80′ )

(2) ¯ (2) ξ22 . J¯2 = E

(2.3.81′ )

The proof of Theorem 2.3.7 will be given below. Remark 2.3.3. Equations (2.3.69) with constant coefficients are well known (e.g., see [1],[189]), but these equations with variable coefficients were apparently first obtained in [83],[84]. Homogenized problems of infinite order specify these equations.

2.3.7

The justification of the asymptotic expansion.

First we prove Lemmas 2.3.2-2.3.5 justifying theorem 2.3.1. Lemma 2.3.2. Let condition A hold and let the matrices Nl (ξ) satisfy the relations Φ∗ Nl = 0, l > 0. Then the elements nkp l of the matrices Nl satisfy the following relations: ˜

δAk +δAp kp nkp nl (SA ξ), l (ξ) = (−1)

δAp = δAp + δp4 ,

A = 2, . . . , s,

(2.3.83)

where k = 1, 2, 3 and p = 1, 2, 3, 4 for s = 3 and k = 1, 2 and p = 1, 2 for s = 2; moreover, the matrices hN l are diagonal. If the sum l + δ1k + δ1p is odd and condition B is satisfied, then the relations nkp l = 0 and qp mq qp mq ∂nl mk + ai1 nl−1 = 0, i = 2, 3, (2.3.84) ϕ aij ∂ξξj

hold (here and below summation over repeated indices from 1 to s is implied).

83


If the sum δ2i + δ2k + δ2p or the sum δ3i + δ3k + δ3p is odd, then for the averages we have ∂nqp mq qp l + a n = 0, i = 2, 3. (2.3.85) ϕmk amq i1 ij l−1 ∂ξξj

The proof of Lemma 2.3.2 is similar to that of [16], Theorems 6.3.3 and 6.3.4: applying the identity δAi +δAj +δAk +δAl kl akl aij (ξ), ij (SA ξ) = (−1)

A = 2, 3, ˜

and substituting into (2.3.11)-(2.3.14)(by induction on l),we see that (−1)δAk +δAp is a solution. Thus, nkp l (SA ξ) ∂npr ∂ ˜ δAp +δÃr l (SA ξ) (−1) = (−1)δAi +δAj +δAp +δAr +δAi +δAj +δAk +δAp akp (ξ) ∂ξξj ∂ξi ij

∂npr ∂npr ∂ kp kp δAk +δÃr ∂ l l (η)

, = (−1) aij (η) × aij (η)

∂ηj η=SA ξ ∂ηj ∂ηi ∂ηi η=SA ξ

∂ kp ˜ δAp +δÃr a (ξ)npr = (−1)δAi +δAk +δAp +δAp +δAr +δAi l−1 (SA ξ)(−1) ∂ξi i1

kp

∂ kp pr pr δAk +δÃr ∂

a (η)nl−1 (η)

, a (η)nl−1 (η) η=S ξ = (−1) × A ∂ηi i1 ∂ηi i1 η=SA ξ (2.3.86) ∂ pr kp δAp +δÃr δAk +δAp +δAj +δAj +δAp +δÃr a1j (ξ) n (SA ξ)(−1) = (−1) ∂ξξj l−1

∂ pr ∂ pr kp δAk +δÃr

n (η) , × akp (η) a n (η) = (−1) (η) 1j 1j l−1 l−1

∂ηj ∂ηj η=SA ξ η=SA ξ (2.3.87)

kp pr pr δAp +δÃr δAk +δÃr kp

a11 (ξ)nl−2 (SA ξ)(−1) = (−1) a11 (η)nl−2 (η) η=S ξ . (2.3.88) A

Finally,

=

ϕ

mq

ϕ

mq

pr mp pr mq ∂nl−1 (SA ξ) δAp +δÃr (ξ) a1j + a11 nl−2 (−1) ∂ξξj δAm +δÃq

(SA ξ)(−1)

δAm +δÃr

(−1)

˜

∂npr mp pr mp l−1

+a11 (η)nl−2 (η) a1j (η) ∂ηj η=SA ξ

˜

= (−1)δAq +δAr hqr l ,

(2.3.89) pr ∂nl−1 (SA ξ) mp pr δAp +δÃr δAp +δÃr (−1) + a n (S ξ)(−1) ϕkq (ξ) ϕmq amp A 11 l−2 1j ∂ξξj ˜

˜

˜

˜ ˜

δAk +δAr = (−1)δAk +δAq ϕkq (SA ξ)(−1)δAq +δAr hqr ϕ(SA ξ)hqr l = (−1) l , (2.3.90)

84


N mq where hqr are entries of the matrix Φ. l are entries of the matrices hl and ϕ By induction on l, from relations (2.3.86)–(2.3.90) we can readily derive that if Nl are solutions to problems (2.3.11)–(2.3.14), then matrices with entries ˜ (−1)δAk +δAp nkp l (SA ξ) are also solutions and for l > 0 both matrices satisfy the relations ˜

Φ∗ Nl = 0,

˜

˜

ϕmk (ξ)(−1)δAm +δAr nl (SA ξ) = (−1)δAk +δAr ϕmk (η)nl (η) = 0.

Since a solution of problem (2.3.11)-(2.3.14) orthogonal to rigid-body disδÃq +δÃr placements is unique, we have (2.3.83). This and (2.3.89) imply hqr l = (−1) N hqr l , i.e., the matrices hl are diagonal. By analogy with (2.3.89), we have qp qp mq qp mq qp mq ∂nl mq ∂nl δAi +δÃk +δÃp mk mk +ai1 nl−1 ; +ai1 nl−1 = (−1) ϕ aij ϕ aij ∂ξξj ∂ξξj

this implies (2.3.85). Similarly, by induction on l, we can prove that if condition B is satisfied, then the matrix (−1)l+δ1p +δ1r npr l (ξ) is a solution to problem (2.3.11)–(2.3.14), l+δ1p +δ1r pr and this implies relations (2.3.84), (2.3.85) and npr nl (ξ). l (ξ) = (−1) qq qq l qq Hence, hl = (−1) hl , hl = 0 for all odd l. This completes the proof of the lemma. qq Lemma 2.3.3. Let condition A be satisfied. Then h11 2 > 0, h2 = 0, qq 44 q = 2, . . . , s, and h2 > 0 for s = 3; moreover, h3 = 0, q = 2, . . . , s. Proof. Let us obtain a new representation for the matrix s ∂N N1 N ∗ + A11 Φ . h2 = Φ A1j ∂ξξj j=1

We have hN 2

=

Φ

∗

s j=1

A1j

∂N N1 ∂ (ξ1 Φ) + ∂ξξj ∂ξξj

s ∂ = Φ∗ A1j (N N1 + ξ1 Φ) . ∂ξξj j=1

Since

∗ s ∂(Φξ1 ) ∂ ∗ = (Aj1 Φ)∗ = Φ∗ A1j , (Φ ξ1 )Amj = Ajm ∂ξm ∂ξm m=1 m=1 s

we obtain hN 2 =

s ∂ ∂ (N N1 + ξ1 Φ) . (Φ∗ ξ1 )Amj ∂ξξj ∂ξm m,j=1

(2.3.91)

∂ ∂ N1∗ Amj (N N1 + ξ1 Φ) . ∂ξm ∂ξξj m,j=1

(2.3.92)

On the other hand, it follows from the integral identity for problem (2.3.11)– (2.3.14) with the test matrix-valued function N1 that 0=

s


85

Summing (2.3.91) and (2.3.92), we obtain the following representation for hN 2 : hN 2 =

∂ ∂ (N N1 + ξ1 Φ) . (N N1 + ξ1 Φ)∗ Amj ∂ξξj ∂ξm m,j=1

s

N Now let us prove that the elements hkl 2 of the matrix h2 satisfy the relations 22 44 > 0 and h2 = 0 and if s = 3, then we also have h33 2 = 0 and h2 > 0. Denote the matrix elements by means of a pair of superscripts. Then we have 1 ∂ ∂ ∂ (N N1 + ξ1 Φ)kl (N N1 + ξ1 Φ)kl akq (N N1 + ξ1 Φ)ql = hll 2 = mj 2 ∂ξm ∂ξm ∂ξξj ∂ (N N1 + ξ1 Φ)ml + ∂ξk ∂ ∂ kq 1 (N N1 + ξ1 Φ)jl (N N1 + ξ1 Φ)ql + ×amj ∂ξq 2 ∂ξξj s 2 ∂ κ ∂ (N N1 + ξ1 Φ)jl . (N N1 + ξ1 Φ)ql + ≥ ∂ξq 4 j,q=1 ∂ξξj

h11 2

Moreover,

2 ∂ (N N1 + ξ1 Φ)11 ∂ξ1 2 ∂ (N N1 + ξ1 Φ)11 = κ > 0, ≥κ ∂ξ1 h11 2 ≥κ

because (∂/∂ξ1 )N N1 = 0 and N1 is 1-periodic with respect to ξ1 . Moreover,

2 ∂N N114 2 ∂ κ ∂ (N N1 + ξ1 Φ)14 +κ (N N1 + ξ1 Φ)r4 + ∂ξ1 ∂ξr ∂ξ1 4 r=2 s

h44 2 ≥

∂n14 2 ∂ 14 2 κ ∂ r4 1 ≥ 0. n1 +κ n1 + ϕr4 + ∂ξ1 ∂ξr ∂ξ1 4 r=2 s

=

44 Let us prove that h44 2 = 0. Assume the contrary, i.e., let h2 = 0. Then the 14 last estimate implies ∂n14 /∂ξ = 0, and n does not depend on ξ1 , moreover, 1 1 1 ∂ ∂ 14 2 r4 n = 0. nr4 + 1 +ϕ ∂ξr 1 ∂ξ1

Hence, 2 ∂nr4 ∂n14 2 ∂nr4 ∂ ∂n14 1 1 1 1 = 0. + ϕr4 + ϕr4 + 2 nr4 +2 1 ∂ξr ∂ξ1 ∂ξr ∂ξ1 ∂ξ1

The second integral and the third integral are equal to zero, because n14 1 does not depend on ξ1 ; hence, n14 1 satisfies the system of equations −aξ3 +

∂n14 1 = 0, ∂ξ2

aξ2 +

∂n14 1 = 0, ∂ξ3

86


whence

∂ 2 n14 1 = a, ∂ξ3 ∂ξ2

∂ 2 n14 1 = −a. ∂ξ2 ∂ξ3

This contradiction proves that h44 2 > 0. The constants hqq vanish for q = 2, . . . , s, because the direct substitution 2 into (2.3.11)–(2.3.13) for l = 2 proves that the second (the third, . . . , the s−th) (2) (s) column, (N N2 , . . . , N2 ), of the matrix N2 satisfies N2q = −(ξq , 0, . . . , 0)∗ ; hence, the corresponding columns of the matrix N2 + ξ1 Φ represent the rigid-body displacements. Now let us prove that hqq 3 = 0 for q = 2, . . . , s. We have p2 p2 p2 p2 ∂ξ2 2p p2 1p ∂n2 1p p2 1p ∂n2 2p ∂n2 +a1p . +a n = a +a n = a h22 = a 3 m1 n1 11 1 21 1 2j mj 1j ∂ξξj ∂ξm ∂ξξj ∂ξξj

Applying the integral identity for problem (2.3.11)–(2.3.14) with l = 2, we transform the last integral to the following form (h12 2 = 0): p2 ∂ ∂np2 1p p2 1p p2 1p ∂n1 12 2 − a n ξ2 + a n ξ = − h − a a1p − 2 2 11 0 m1 1 1j ∂ξξj ∂ξm mj ∂ξξj

=

a1p 1j

∂np2 p2 1p ∂ p2 1 (n + ξ δ ) ξ2 = 0, + a1p n ξ = a 1 p 2 2 11 0 1j ∂ξξj 1 ∂ξξj

since the second column of the matrix N1 + ξ1 Φ is the rigid-body displacement. For the case s = 3, we similarly obtain h33 3 = 0. This proves the lemma. Let the elasticity operator −

s ∂ x ∂u , Aij µ ∂xj ∂xi i,j=1

x ∈ Uµ ,

(2.3.93)

and the derivation with respect to the co-normal −

s x ∂u ∂u , =− ni Aij µ ∂xj ∂ν i,j=1

x ∈ ∂U Uµ ,

(2.3.94)

be given and let u be a vector function from Hb1−per (U Uµ ); this space is the completion (by the H 1 (C Cµ )−norm) of the space of b-periodic in x1 differentiable in U¯µ vector-valued functions. Assume that ∫Cµ Φ∗ u dx = 0. Lemma 2.3.4. The function u satisfies the following inequality (for sufficiently small µ): ∂u ∂u dx ≥ c µ4 ∇u2L2 (Cµ ) + µ2 u2L2 (Cµ ) , c > 0. (2.3.95) , Aij ∂xj ∂xi Cµ

Proof.

87


Let us perform the following changes of variables in formulas (2.3.93) and (2.3.94): 1) x1 = ξ1 , x2 = ξ2 µ, . . . , xs = ξs µ, 2) u1 = w1 , u2 = w2 /µ, . . . , us = ws /µ, 3) multiply the components of the vector (2.3.93), from the second to the s−th one, and the first component of the vector (2.3.94) by µ−1 and multiply the components of the vector (2.3.94) from the second to the s−th one, by µ−2 . Then formulas (2.3.93) and (2.3.94) take the form −

s ∂w ∂ , Bij (ξ) ∂ξξj ∂ξi i,j=1

−

s

ξ ∈ Q0 = {ξ1 ∈ (0, b), ξ ′ ∈ β},

ni Bij (ξ)

i,j=1

∂w , ∂ξξj

ξ ′ ∈ ∂β,

(2.3.96)

(2.3.97)

where the entries bkl ij of the matrices Bij are kl bkl ij (ξ) = aij

Clearly,

ξ

1

µ

, ξ′

1 δi2 +···+δis +δj2 +···+δjs +δk2 +···+δks +δl2 +···+δls

µ

.

il lk bkl ij = bkj = bji

and for any symmetric matrix with elements ηik (ηik = ηki ) the following inequality holds: s k l bkl ij ηi ηj = i,j,k,l=1

=

s

l −(δj2 +···+δjs +δl2 +···+δls ) k −(δi2 +···+δis +δk2 +···+δks ) akl ηi µ ij ηj µ

i,j,k,l=1

≥κ

s

ηjl µ−(δj2 +···+δjs +δl2 +···+δls )

j,l=1

Moreover, we have

2

Φ∗ w dξ = 0.

≥κ

s

(ηjl )2 .

j,l=1

(2.3.98)

Q0

The following inequality holds for any function from H 1 (Q0 ) that satisfies relation (2.3.98) (see [55]) s ∂w ∂w dξ ≥ κcw2W 1 (Q0 ) , Bij , 2 ∂ξξj ∂ξi i,j=1 Q0

c > 0, where c is the constant of Korn’s inequality for Q0 . Since we have s s ∂w ∂w l k dξ = µ−(s−1) , akl Bij ij ej (u)ei (u) dx, ∂ξ ξ ∂ξ i j C Q µ 0 i,j=1 i,j,k,l=1

88


where elj (u) =

and since

∂ul 1 ∂uj , + ∂xj 2 ∂xl

w2H 1 (Q0 ) ≥ µ−s+5 ∇u2L2 (Cµ ) + µ−s+3 u2L2 (Cµ ) ,

we obtain the inequality of Lemma 2.3.4. Lemma 2.3.5. The diagonal elements of the matrix hN 4 , from the second to the s−th one, are negative. Proof. Let the function u(K) µ (x) =

K+1

µl Nl

l=0

x ∂ l V (x ) 1 µ ∂xl1

be specified in the space Uµ , where Nl (ξ) are the matrices constructed above, V (x1 ) ∈ C ∞ (IIR) is a b-periodic d-dimensional vector that does not depend on µ, and K is sufficiently large. One can verify by direct substitution that K+1 s (K) ∂lV ∂ x ∂uµ + θk , =Φ µl−2 hN Aij l µ ∂xj ∂xi ∂xl1 i,j=1 l=2

where

s ∂ K+2 V ∂ (Ai1 NK +1 ) K+2 , ∂ξi ∂x1 i=2 K+2 s V ∂ ∂N NK +1 ∂ + A11 NK (A11 NK +1 ) + A1j θ¯k = µK K+2 ∂ξξj ∂ξ1 ∂x1 j=1

θk = θ¯k + θ¯k ,

θ¯k = µK

+µK+1 A11 NK +1 s

(K)

ni Aij

i,j=1

∂uµ ∂xj

=

s i=2

∂ K+3 V ; ∂xK+3 1

µK+1 ni Ai1 NK +1

∂ K+2 V , ∂xK+2 1

Uµ . x ∈ ∂U

A similar discrepancy arises on the discontinuity surfaces for the coefficients Σµ . On the other hand s s (K) (K) (K) ∂uµ ∂uµ ∂uµ (K) ∂ dx , , uµ dx = − Aij Aij ∂xi ∂xj ∂xj ∂xi Cµ i,j=1 Cµ i,j=1

+

∂Cµ

s

i,j=1

ni Aij

s (K) (K) ∂uµ (K) ∂uµ , u ds, , u(K) ds + n A i ij µ µ ∂xj ∂xj Σµ i,j=1

whence the integral I has the form K+1 ∂ l V (K) , u dx Φ I =− µl−2 hN l ∂xl1 Cµ l=2


=

s

Aij

Cµ i,j=1

(K) (K) ∂uµ ∂uµ dx + O(µK+s−1 ). , ∂xi ∂xj

89

(2.3.99)

Since Φ∗ Nl = δl0 , by analogy with the proof of [16], Lemma 4.2.1, we obtain K+1 l l−2 N ∂ V , V dx + O(µK ) I =− Φ µ hl l ∂x Cµ 1 l=2 s−1

= −µ

K+1

l−2

µ

mes β

b

0

l=2

∂lV , V dx1 + O(µK+s−1 ). l ∂x1

hN l

Let the first and the fourth (for s = 3) components of the vector V be zero. Then (K > 3) b ∂4V , V dx1 + O(µs+2 ) hN I = −µs+1 mes β 4 4 ∂x1 0

and it follows from (2.3.99) that s+1

−µ

Cµ

s

i,j=1

Aij

mes β

b

0

(K) ∂uµ

∂xj

,

hN 4

(K)

∂uµ ∂xi

∂4V , V dx1 = 4 ∂x1

dx + O(µs+2 ).

(2.3.100)

It follows from Lemma 2.3.4 that the right-hand side of this equality is bounded (K) from below by c1 µ2 uµ 2L2 (Uµ ) ; in its turn, for small µ, this expression is bounded from below by c1 µs+1 c1 µ2 mes β V 2L2 (0,b) . ΦV 2L2 (Uµ ) = 2 2 Here constant c1 is independent of µ. By taking the second component of the vector V , for example, in the form sin(2πx1 /b) and by setting the other components equal to zero, we see from equation (2.3.100) that h22 4 < 0. Similarly for s = 3 we can prove that h33 4 < 0. The statements of Theorem 2.3.1 follow from Lemmas 2.3.2–2.3.5. Lemma 2.3.6. If condition A for problem (1)–(4) is satisfied, then the r r matrices hN and hN have the following structure: if s = 2, then 0 1 ⎛ ⎞ 1 0 r ⎝0 1⎠ hN 0 = 0 0 r is equal to (0, 1); and the third row of hN 1 ⎛ 1 0 ⎜0 1 ⎜ ⎜0 0 Nr h0 = ⎜ ⎜0 0 ⎜ ⎝0 0 0 0

if s = 3, then ⎞ 0 0 0 0⎟ ⎟ 1 0⎟ ⎟ 0 1⎟ ⎟ 0 0⎠ 0 0

90


r and the fifth row and the sixth row of the matrix hN are equal to (0, 1, 0, 0) and 1 to (0, 0, 1, 0), respectively. Proof. Since N0 = Φ, the trivial matrices N0r = 0 are solutions of boundary layer r problems (2.3.30) and (2.3.31), and hN indeed has the above structure; so, 0

(j)

The columns N1 equal to

r ′ ξ ′ )hN Φ(0, 0 = Φ(ξ ).

of the matrix N1 , from the second to the sth one, are (j)

N1

= (−ξξj , 0, . . . , 0)∗ ,

j = 2, . . . , s r(j)

(see the proof of Lemma 2.3.3). Then the columns N1 of the matrices N1r , from the second to the s−th one, are solutions of the homogeneous equations r(j) Lξξ N1 = 0 with boundary conditions r(j)

∂N N1 ∂ννξ

= 0 for ξ ′ ∈ ∂β

r(j)

and N1

N r(j)

(j)

N r(j)

ξ ′ )h (0, ξ ′ ) = −N N1 (0, ξ ′ ) + Φ(0, 1

,

r where h1 is the j−th column of the matrix hN 1 . Setting the third element N r(j) of h1 equal to 1 for s = 2 and the (j + 3)−th element equal to 1 for s = 3 r(j) and setting the other elements equal to zero, we see that N1 = 0 is a solution of the boundary layer problem. To complete the proof, it remains to establish that the first (for s = 3, the last) element of the last (for s = 3, the next to the last) row of the matrix r is equal to zero. To do this, it suffices to prove that for s = 3, the first hN 1 and the fourth columns of a solution to problem (2.3.37) stabilize to the rigidbody displacement orthogonal to the vectors (−ξ2 , ξ1 , 0) and (−ξ3 , 0, ξ1 ) and for s = 2, the first column of solution to problem (2.3.37) stabilizes to a constant. 0 to problem In its turn, this assertion follows from the fact that the solution N 1 (2.3.37) satisfies (under condition A) relations similar to equations (2.3.83): ˜

δAk +δAp kp n kp n 1 (SA ξ), 1 (ξ) = (−1)

A = 2, . . . , s.

(2.3.101)

Indeed, it follows from (2.3.101) that n 11 14 1 (ξ) is an even function (and n 1 (ξ) is an odd function for s = 3) of ξ2 , . . . , ξs , hence, they cannot stabilize to a rigid-body displacement with the first component of the form c0 + c2 ξ2 + · · · + cs ξs whose 0 stabilizes coefficients c2 , . . . , cs satisfy |c2 | + · · · + |cs | = 0. This means that N 1 N0 N0 to the rigid-body displacement (−Φh1 ), where the matrix h1 has the zero first element in the last row for s = 2 and the first and the last zero elements in the two last rows for s = 3. The proof of relations (2.3.101) is perfectly similar to that of Lemma 2.3.2. This completes the proof of Lemma 2.3.6. The proof of Theorems 2.3.2, 2.3.3, and 2.3.5 is perfectly similar to the proof of Theorems 4 and 5 in [130] and of the theorems (of Phr¨ a¨gmen-Lindelöf type) 8.1 and 8.3 of Chapter 1 in [129].

91


Lemma 2.3.7. Let condition A be satisfied. Then the solutions of problems (30′ ) satisfy the following relations 0 pq nl (ξ)

˜

= (−1)δAp +δAq 0 npq l (SA ξ),

χqr l =

A = 2, . . . , s,

(2.3.102)

∂ 0 npr l−1 0 pr + akp ϕ mq akp 11 nl−2 dξ 1j ∂ξξj (0,+∞)×β ˜ ˜

˜

= (−1)δAq +δAr χqr A = 2, . . . , s. l , ⎧ ⎨ δAq for q = 1, . . . , d, δAq = δA,q−3 for q = 5, 6, s = 3, ⎩ 1 for q = 3, s = 2.

(2.3.103)

0 Here 0 npq l are the entries of the matrices Nl (ξ). The proof of relations (2.3.102) is similar to that of Lemma 2.3.2 (by substitution). Relation (2.3.103) follows from equations (2.3.102): ˜ ˜ ˜ qr χl = (−1)δAk +δAq ϕ kq (η)(−1)δAk +δAp +δAp +δAr (0,+∞)×β

∂ 0 npr ˜ ˜ kp kp l−1 0 pr dξ = (−1)δAq +δAr χqr + a11 (η) nl−2

× a1j (η) l . ∂ηj η=SA ξ

0 It follows from Lemmas 2.3.2 and 2.3.7 that nonzero elements of hN can l occupy only the following positions marked by stars: for s = 3 we have ⎛ ⎞ ∗ 0 0 0 ⎜0 ∗ 0 0⎟ ⎜ ⎟ ⎜0 0 ∗ 0⎟ ⎜ ⎟ ⎜0 0 0 ∗⎟ , ⎜ ⎟ ⎝0 ∗ 0 0⎠ 0 0 ∗ 0

for s = 2 we have

⎛ ∗ ⎝0 0

⎞ 0 ∗⎠ . ∗

Lemma 2.3.8. For problem (2.3.1)-(2.3.4), (2.3.3’) we have 0 qq h1

= −hqq 2 0 32 h2 0 pq hl

for

q = 1, . . . , d,

= ξ22 −1 h22 4

for

s = 2,

0 q+3,q h2

= ξq2 −1 hqq 4

0 qq h3 0 hN l .

= −hqq 4

for

for

q = 2, 3, s = 3,

q = 2, . . . , s.

are the entries of the matrices Here Proof. Since T1N 0 = 0 and the second column (and the third column for s = 3) of ¯ qq and h ¯ q+3,q , over the matrix T2N 0 is zero, the integral in formulas (2.3.45) for h 1 2

92


the set (0, +∞) × β, vanishes. Let us prove that this integral also vanishes in ¯ qq . Let q = 2 and s = 3 (the proof for q = 3 and s = 2 is formulas (2.3.45) for h 3 similar). We have: 0 p2 ∂ξ2 ∂ 0 np2 1p 0 p2 2p 0 p2 1p ∂ n2 2 dξ +a n dξ = a +a n a2p m1 1 11 1 mj 1j ∂ξm ∂ξξj ∂ξξj (0,+∞)×β (0,+∞)×β

0 p2

∂ ∂ 0 np2 1p ∂ n2 1p 0 p2 1p 0 p2 2 a +a n ξ2 dξ +a n ξ − = − a1p 2 m1 1 11 1 mj 1j

∂ξξj ∂ξξj (0,+∞)×β ∂ξm ξ1 =0

=

∂np2 2 a1p 1j ∂ξξj

+

p2 + a1p n 11 1

(0,+∞)×β

a1p 1j

ξ2

ξ1 =0

+

d¯ ( p=1

ϕ 1p (0, ξ ′ ) 0 hp2 2 ξ2

∂ 0 np2 0 p2 1 + a1p n ξ2 dξ. 11 0 ∂ξξj

)

0 p2 The last integral vanishes because 0 np2 1 = n0 = 0. By equations (2.3.45), we have d¯ ( 1p ) 2 0 52 ϕ (0, ξ ′ ) 0 hp2 2 ξ2 = −ξ2 h2 = p=1

=

ξ22 ξ22 −1

q2 q2 1q q2 lq q2 1q ∂n2 lq ∂n2 + a11 n1 ξ2 , = − a1j + a11 n1 ϕ (0, ξ ) a1j ∂ξξj ∂ξξj l5

′

whence the above integral is zero. qq qq 0 qq Now the equations 0 hqq 1 = −h2 for q = 1, . . . , d, and h3 = −h4 follow from the fact that relations (2.3.45) and (2.3.14) give the same expressions for these quantities. Finally, 2 −1 −0 h52 2 = −ξ2

a1q 1j

q2 ∂ ∂nq2 1q q2 1q ∂n3 q2 2 −1 2 ξ2 +a n a +a1q n ξ = ξ 2 2 m1 2 11 1 ∂ξm mj ∂ξξj ∂ξξj

∂nq2 ∂nq2 q2 q2 3 3 +a2q = −ξ22 −1 h22 = −ξ22 −1 a1q = −ξ22 −1 a2q +a1q 4 . 11 n2 21 n2 2j 1j ∂ξξj ∂ξξj

0 32 The statements concerning 0 h63 2 and h2 for s = 2 can be proved similarly. This completes the proof of Lemma 2.3.8.

The statement of Theorem 2.3.4 follows from the statements of Lemmas 2.3.7 and 2.3.8, taking into account the homogeneity of equations (2.3.58). The proof of Theorem 2.3.6 can be given similarly to that of Lemmas 2.3.6–2.3.8. Lemma 2.3.9. A solution of the equation P u = f in the domain Cµ with boundary conditions (2.3.2),(2.3.3) and with the right-hand side f ∈ L2 (C Cµ ) satisfies the a priori estimate uH 1 (Cµ ) ≤ cµ−2 f L2 (Cµ ) .

93


Proof. By analogy with Lemma 2.3.4, we obtain estimate (2.3.95). Furthermore, it follows from the integral identity that ∂u ∂u dx = − (f, u) dx ≤ , Aij ∂xj ∂xi Cµ Cµ

≤ f L2 (Cµ ) uL2 (Cµ ) ≤ f L2 (Cµ ) uH 1 (Cµ ) . This, together with inequality (2.3.95), implies the desired estimate. Lemma 2.3.9 is used to obtain the estimate for the proximity of the exact solution of problem (2.3.1)–(2.3.4) and the asymptotic expansion u(∞) . Namely, as it was done in section 2.2, substitute the (K + 1)−th partial sum u(K) of the series u(∞) into equations (2.3.1)–(2.3.4). Grouping the terms in the same way as in the series itself, we see that the residual on the right-hand side of the equation can be estimated by O(µK−1 ). A similar residual in the boundary conditions is transferred to the right-hand side by substitution (see section 2.2). Then we can apply the a priori estimate of Lemma 2.3.9 and obtain u − u(K) H 1 (Cµ ) = O(µK ) mes Cµ .

The solvability of the ordinary differential equations for ωj is clear. Thus we obtain

Theorem 2.3.8. Let K ∈ {0, 1, 2, ...}. Denote χ a function from C (K+2) ([0, b]) such that χ(t) = 1 when t ∈ [0, b/3], χ(t) = 0 when t ∈ [2b/3, b]. Consider a function (K)

0(K)

u(K) (x) = uB (x) + uP

1(K)

(x)χ(x1 ) + uP

˜ (x)χ(b − x1 ) + Φ(0,

x′ )ρ(x1 ), µ

where (K)

uB

=

K+1

µl Nl

l=0

0(K)

uP

=

K+1

x dl ψ(x ) x dl ω (K) (x ) K−1 1 1 , + µl+2 Ml l l µ µ dx dx1 1 l=0

µl Nl0 (ξ)

l=0

1(K)

uP

=

K+1 l=0

µl Nl1 (ξ)

K−1 dl ψ

dl ω (K) , + µl+2 Ml0 (ξ) l l dx1 ξ=x/µ dx1 l=0

K−1 dl ψ dl ω (K) l+2 1 |ξ =(x −b)/µξ′ =x′ /µ , + µ M (ξ) l dxl1 1 1 dxl1 l=0

ω (K) =

K

µj ωj (x1 );

j=−2

ρ(x1 ) = (1 − x1 /b)q0 + (x1 /b)q1 ,

94


qr =

2K+1

µl

l=K+1

r hN p

j+p=l,−2≤j≤K,0≤p≤K+1

K−1 ψ dp ω j K+1 M r d hK−1 K−1 |x1 =rb , r = 1, 2, p +µ dx1 dx1

Nl , Nl0 , Nl1 , Ml , Ml0 , Ml1 , ωj are defined in subsection 2.3. Then the estimate holds u(K) − u(H 1 (Cµ ))s = O(µK ) mes Cµ .

Proof repeats the proof of Theorem 2.2.3 with the following corrections. 1. The products of type Amj

are replaced by inner products

Amj

x ∂u(K) ∂φ B µ ∂xj ∂xm

x ∂u(K) B

∂xj

µ

,

∂φ . ∂xm

N 2. At the second stage of the proof, the integral J˜B (φ) is presented in the form K+1 s ∂ dl ω (K)∗ (x1 ) N (φ |x=µξ − J˜B (φ) = µs+l−2 ABN (ξ) : ml ∂ξm dxl1 µ−1 Cµ m=1 l=0

− ABN 1,l−1 (ξ) : s+K

−µ

µ−1 Cµ

dl ω (K)∗ (x ) 1 |x=µξ dξ − φ dxl1

ABN 1,K+1 (ξ) :

dK+2 ω (K)∗ (x ) 1 |x=µξ dξ, φ dxK+2 1

where : denotes the element-wise multiplication of matrices, i.e. if A and B are two matrices of the same dimension with the elements aij and bij respectively then A : B is the matrix with the elements aij bij (no summation!). 3. At the fifth stage of the proof, I(φ) =

Cµ

−F˜ ψ − Φ(

F ψ − Φ(

x′ ) µ

K−1 l=1

K+1 x′ l−2 N dl ω (K) − µ hl ) µ dxl1

µl hM l

l=0

dl ψ φ(x) dx + ∆3 (φ). dxl1

4. The a priori estimate constant of lemma 2.3.9 depends on µ and therefore (2.2.40) transforms into µ4 κu − u(K) 2H 1 (Cµ ) ≤ I(u − u(K) ) =

95


=

s

Cµ m,j=1

Amj

x ∂(u − u(K) ) ∂(u − u(K) )

µ

∂xm

∂xj

≤ c10 µK u − u(K) H 1 (Cµ )

dx ≤

mes Cµ ,

(2.3.40′ )

with a constant c10 > 0 independent of µ. So, u − u(K) H 1 (Cµ ) ≤ c10 /κµK−4 mes Cµ ;

and therefore

u − u(K+4) H 1 (Cµ ) ≤ c10 /κµK By definition, u(K) − u(K+4) H 1 (Cµ ) = O(µK

Thus, by triangle inequality, u − u(K) H 1 (Cµ ) = O(µK

mes Cµ .

mes Cµ ).

mes Cµ ).

This completes the proof of the theorem. Similarly we can prove the analogous theorems for problems (2.3.1), (2.3.2), (2.3.3’) and (2.3.48),(2.3.4)). Proof of Theorem 2.3.7. Substituting np1 1 = (1 − δp1 )

(−λ) (−λ) ξp ξp for p = 2 and np1 1 = (1 − δp1 ) 2(M + λ) 2M + λ

for s = 3 into (2.3.11)–(2.3.13), we directly verify that these functions satisfy problem (2.3.11)–(2.3.13) for l = 1. Then * s p1 (λ+2M )2 −λ2 for s = 2, 1p ∂n1 11 (s) 11 λ+2M ¯ = + a11 = E h2 = a1j M (3 λ+2M ) ∂ξ ξ for s = 3. j λ+M j,p=2 ′ The function n14 1 is a solution Y (ξ ) of the Laplace equation in the domain β with the boundary condition ∂Y /∂n − aξ3 n2 + aξ2 n3 = 0 on the boundary ∂β. Then ∂n114 ∂n114 2 2 + ξ a + ξ a + ξ h44 = Ma − ξ 2 3 2 3 2 ∂ξ3 ∂ξ2 Ma a ξ22 + ξ32 dξ ′ + (−ξ3 n2 + ξ2 n3 )n14 ds . (2.3.104) = 1 mes β β ∂β

96


On the other hand, for each z ∈ H 1 (β), the function n14 1 satisfies the integral identity 14 ∂n114 ∂z ′ ∂n1 ∂z (−ξ3 n2 + ξ2 n3 )z ds. (2.3.105) dξ = −a + ∂ξ3 ∂ξ3 ∂ξ2 ∂ξ2 ∂β β

By setting z = n14 1 and replacing the last integral in (104) according to (2.3.105), we obtain ∂n 14 2 ∂n 14 2 2 2 2 1 1 = − = M a (ξ + ) − h44 ξ 2 2 3 ∂ξ3 ∂ξ2 14 ∂n1 2 ∂n114 2 ≤ M. + =M 1− ∂ξ3 ∂ξ2

Consider problems (2.3.11)–(2.3.14) for the second column (for s = 3, for (2) (3) the third column) of Nl : Nl , Nl . For s = 2 we obtain (2)

N0

= (0, 1)∗ ,

(2)

N1

= (−ξ2 , 0)∗ ,

(2)

N2

=

λ (0, ξ22 − ξ22 )∗ , 2(λ + 2M )

(2)

A12

∂N N3 (2) ¯ (2) (ξ22 − 1/4)/2)∗ ; + A11 N2 = (0, E ∂ξ2

and for s = 3 we have (2)

= (0, 1, 0)∗ ,

(2)

= (−ξ2 , 0, 0)∗ , ∗ ν λ (2) ; N2 = 0, (ξ22 − ξ32 − ξ22 − ξ32 ), νξ2 ξ3 , ν = 2(M + λ) 2 p2 ∂np2 1p p2 ip p2 1p ∂n3 3 + a n + a n = a aip 11 2 i1 2 ij 1j ∂ξξj ∂ξξj ∂ 1p ∂np2 ∂np2 p2 p2 ∂ξi 3 ξi amj 3 + a1p = − + a1p = a1p m1 n2 m1 n2 mj ∂ξξj ∂ξm ∂ξm ∂ξξj ∂np2 p2 12 2 ¯ (3) ξ 2 δi2 , + a1p = a1p 2 11 n1 − h3 ξi = −E 1j ∂ξξj N0

N1

(2)

where the next to last equation follows from the fact that N3 is a solution to (2) problem (11)–(14), and the last equation follows from the explicit form of N2 (2) and N1 . Here summation over repeated indices j and m from 2 to 3 and over the index p from 1 to 3 is implied. Similarly one can prove that

aip 1j

∂np3 p3 3 ¯ (3) ξ 2 δi3 . + aip n = −E 3 11 2 ∂ξξj

¯ (s) ξr2 < 0, r = 2, 3, and hrr = 0, r = 2, 3. This completes the Hence hrr 4 = −E 2 proof of Theorem 2.3.7.


97

Remark 2.3.2. Applying the explicit formulae for components of Nl we can prove that in isotropic case any sufficiently smooth d−dimensional vector-valued function ω(x1 ) satisfies the asymptotic equality ∞ l=1

s dl ω ∂N Nl ˜ ∗ (ξ) = + A11 Nl−1 µl−1 Φ A1j ∂ξξj dxl1 j=1

⎞ ⎧⎛ ¯ dω1 + O(µ) E ⎪ ⎪⎜ dx1 ⎪ ⎪ ⎪⎜−µ2 Eξ ¯ 2 d3 ω32 + O(µ3 )⎟ ⎟ f or s = 2, ⎪ 2 dx ⎪ ⎠ ⎪ ⎪⎝ ¯ 2 d2 ω2 1 2 ⎪ ⎪ ⎪ µEξ2 dx21 + O(µ ) ⎪ ⎪ ⎞ ⎛ ⎪ ¯ dω1 + O(µ) ⎪ E ⎪ dx1 ⎨ ⎜ ¯ 2 d3 ω32 + O(µ3 )⎟ = ⎜−µ2 Eξ ⎟ 2 dx1 ⎟ ⎜ ⎪ ⎪ 3 ⎟ 2 ¯ 2 d3 ω 3 ⎜ ⎪ + O(µ ) ⎪⎜−µ Eξ3 dx3 ⎟ ⎪ 1 ⎪ ⎟ f or s = 3. ⎪⎜ ⎪ dω 4 ¯ ⎟ ⎜ ⎪ + O(µ) M ⎪ dx 1 ⎟ ⎪⎜ 2 2 ⎪ d ω ⎜ 2 ⎪ 2 ¯ ⎪⎝ µEξ2 dx2 + O(µ ) ⎟ ⎪ ⎠ ⎪ 1 ⎪ ⎩ µEξ ¯ 2 d2 ω23 + O(µ2 ) 3 dx 1

Denote the main term of this asymptotic expression as Λµ ω, i.e.

⎞ ⎧⎛ ¯ dω1 E ⎪ ⎪⎜ dx1 ⎪ ⎪ ⎪⎜−µ2 Eξ ¯ 2 d3 ω32 ⎟ ⎟ ⎪ 2 dx1 ⎠ f or s = 2, ⎪ ⎝ ⎪ 2 2 ⎪ d ω ⎪ 2 ¯ 2 ⎪ ⎪ µEξ 2 dx1 ⎪ ⎪ ⎞ ⎛ ⎪ ¯ dω1 ⎪ E ⎪ dx1 ⎨ ⎜ ¯ 2 d3 ω32 ⎟ Λµ ω = ⎜−µ2 Eξ 2 dx1 ⎟ ⎟ ⎜ ⎪ ⎪ ⎪⎜−µ2 Eξ ¯ 2 d3 ω33 ⎟ ⎪ 3 ⎟ ⎜ ⎪ dx 1 ⎪ ⎟ f or s = 3. ⎪⎜ ⎪ dω 4 ¯ ⎟ ⎜ ⎪ M ⎪⎜ dx1 ⎟ ⎪ 2 2 ⎪ d ω ⎜ ⎪ 2 ¯ ⎪⎝ µEξ2 dx2 ⎟ ⎪ ⎠ ⎪ 1 ⎪ ⎩ µEξ ¯ 2 d2 ω23 3 dx 1

Remark 2.3.3. We have studied in this section the case when the rod satisfies some symmetry conditions (Condition A). The general case was studied by different techniques in [117,119] and by the techniques of this section in [101] by A. Majd. The main idea is in multiplication of components of the ansatz by some normalizing factors.

98


2.4

Non steady-state conductivity of a rod

2.4.1


Consider a non steady-state conductivity equation (s = 3!) −ρ(x/µ)

3 ∂u ∂ ∂u = Fµ (x, t), Amj (x/µ) + ∂xj ∂t m,j=1 ∂xm

x ∈ Cµ ,

(2.4.1)


u = 0,

x1 = 0

x ∈ ∂U Uµ ,

or x1 = b.

(2.4.2) (2.4.3)

and with junction conditions s

[u] = 0;

i,j=1


(2.5.4)

(as in section 2.3) and with initial conditions

u t=0 = 0.

(2.4.5)

Here Aij (ξ) are functions from section 2.2 and the function ρ(ξ) is 1-periodic with respect to ξ1 , ρ(ξ) ≥ κ > 0, and ρ(ξ) is piecewise smooth in the sense of section 2.2. Here κ is the same as in section 2. Besides, ρ(SA ξ) = ρ(ξ) for A = 2, 3 and x ψ(x1 , t), Fµ (x, t) = F µ

where F is the same as in section 2.2 and ψ(x1 , t) ∈ C ∞ ,

ψ(x1 , t) = 0 for t ≤ θ0 ,

θ0 > 0.

The existence and uniqueness of a solution follows from [85],[54]. This solution belongs to the space H 1,0 (C Cµ × (0, T )) of functions with a norm , T

uH 1,0 (Cµ ×(0,T )) =

2.4.2

sup u(x, t)2L2 (Cµ ) +

t∈(0,T )

0

u(x, t)2H 1 (Cµ ) dt.

Inner expansion

An inner f.a.s. of problem (2.4.1)-(2.4.5) is sought as a series u(∞) =

∞

l+q=0

µl+q Nlq (x/µ)

∂ l+q ω(x1 , t) + ∂xl1 ∂tq

2.4.

99

NON STEADY-STATE CONDUCTIVITY OF A ROD ∞

+

µl+q+2 Mlq (x/µ)

l+q=0

∂ l+q ψ(x1 , t) , ∂xl1 ∂tq

(2.4.6)

where ω is a C ∞ −function and Nlq and Mlq are analogs of the functions Nl and Ml from section 2; Nlq are (1-periodic with respect to ξ1 ) solutions of the problems Lξξ Nlq = −T TlqN (ξ) + hN ξ ′ ∈ β, (2.4.7) lq , 3 ∂N Nlq =− Am1 Nl−1,q nm , ∂ννξ m=2

-

[N Nlq ] Σ = 0,

TlqN (ξ) =

ξ ′ ∈ ∂β,

(2.4.8)

.

3

∂N Nlq

= − A N n m1 l−1,q m , ∂ννξ Σ Σ m=1

(2.4.9)

3 3 ∂ ∂N Nl−1,q + A11 Nl−2,q − ρN Nl−1,q−1 , (Aj1 Nl−1,q ) + A1j ∂ξξj ∂ξξj j=1 j=1

hN lq =

3

∂N Nl−1,q A1j + A11 Nl−2,q − ρN Nl−1,q−1 ∂ξξj j=1

(2.4.10)

,

(2.4.11)

Mlq are solutions of the same problems with Nlq replaced by Mlq . However, M00 is a solution of problem (2.2.15). As in section 2.2, we obtain a homogenized equation −ρ

+

∞

∂2ω ∂ω + + hN 2 ∂x21 ∂t

µl+q−2 hN lq

l+q=3

+

∞

µl+q hM lq

l+q=0

∂ l+q ω + ∂xl1 ∂tq

∂ l+q ψ − F¯ ψµ = 0, ∂xl1 ∂tq

(2.4.13)

where the positive coefficient hN 2 is the same as in section 2. Let us seek f.a.s. ω of (2.4.13) in the form of a regular series in powers of µ: ω=

∞

µj ωj .

(2.4.14)

j=0

Substituting (2.4.14) into (2.4.13) we obtain a chain of parabolic equations −ρ

∂ 2 ωj ∂ωj = fj (x1 , t), + hN 2 ∂x21 ∂t

(2.4.15)

where right hand sides fj are some linear combinations of derivatives of functions ωr with r < j and f0 = F¯ ψ.

100


2.4.3

Boundary layer corrector

The boundary layer corrector can be constructed similarly to that in section 2.2. The f.a.s. to problem (2.4.1)–(2.4.5) is sought in the form, u(∞) = uB + u0P + u1P ,

(2.4.16)

where uB is defined by equation (2.4.6) and ∞

u0P =

0 µl+q Nlq (x/µ)

l+q=0

+

∞

0 µl+q+2 Mlq (x/µ)

l+q=0

∞

u1P =

∂ l+q ψ(x1 , t)

,

ξ=x/µ ∂xl1 ∂tq

1 µl+q Nlq (x/µ)

l+q=0

+

∞


l+q=0

∂ l+q ω(x1 , t) + ∂xl1 ∂tq

∂ l+q ω(x1 , t) + ∂xl1 ∂tq

∂ l+q ψ(x1 , t) |ξ1 =(x1 −b)/µξ′ =x′ /µ ., ∂xl1 ∂tq

(2.5.17)

r r and Nlq , Mlq are exponentially decaying in ξ1 , 1-periodic in all variables except of ξ1 functions. They are found from the chain of boundary value problems similar to that of section 2.2.4, but right hand sides of the boundary layer problems contain the supplementary term −ρN Nlr−1,q−1 , r = 0, 1 (or respectively r −ρM Ml−1,q−1 , r = 0, 1).

2.4.4

Justification

Theorem 2.4.1. Let K ∈ {0, 1, 2, ...}. Let χ be a function from section 3. Consider a function (K)

0(K)

u(K) (x, t) = uB (x, t) + uP

1(K)

(x, t)χ(x1 ) + uP

(x, t)χ(b − x1 ) + ˜(x1 , t),

where (K)

uB

=

K+1

µl+q Nlq

l+q=0

+

K−1

µl+q+2 Mlq

l+q=0

u0P =

K+1

l+q=0

x ∂ l+q ω(x , t) 1 + µ ∂xl1 ∂tq

x ∂ l+q ψ(x , t) 1 , µ ∂xl1 ∂tq

0 µl+q Nlq (x/µ)

∂ l+q ω(x1 , t) + ∂xl1 ∂tq

2.4.

101

NON STEADY-STATE CONDUCTIVITY OF A ROD K−1

+


l+q=0

u1P =

K+1

1 µl+q Nlq (x/µ)

l+q=0

+

K−1

∂ l+q ψ(x1 , t)

,



l+q=0

∂ l+q ω(x1 , t) + ∂xl1 ∂tq

∂ l+q ψ(x1 , t) |ξ1 =(x1 −b)/µξ′ =x′ /µ ., ∂xl1 ∂tq

ω (K) =

K

µj ωj (x1 );

j=0

ρ˜(x1 , t) = (1 − x1 /b)q0 (t) + (x1 /b)q1 (t),

qr (t) =

2K+1

l+q=K+1

+µK+1

µl+q−2

r hN pα

j+p+α=l,0≤j≤K,0≤p+α≤K+1

r hM lq

l+q=K−1

∂ p+α ωj + ∂xp1 ∂tα

∂ K−1 ψ |x1 =rb , r = 1, 2, ∂xl1 ∂tq

0 1 0 1 Nlq , Nlq , Nlq , Mlq , Mlq , Mlq , ωj are defined in subsections 2.2.2, 2.2.3. Then the estimate holds

u(K) − uH 1,0 (Cµ ×(0,T )) = O(µK ) mes Cµ .

(2.4.18)

Justification of the asymptotic is also similar to that in section 2.2. Namely, it is well known that applying the improving smoothness technique [85],[86],[54] (K)

) belongs to H 1 (C Cµ ) for all t ∈ (0, T ). So as in it can be shown that ∂(u−u ∂t section 2.2 we estimate the discrepancy functional

It (φ) =

+

s

Cµ m,j=1

x ∂(u − u(K) ) φdx + ρ ∂t µ Cµ

Amj

x ∂(u − u(K) ) ∂φ dx, ∂xm ∂xj µ

(2.4.19)

where for all t ∈ (0, T ), φ ∈ H 1 (C Cµ ), φ = 0 for x1 = 0 or b. The second integral is developed in the same way as in section 2.2, while the contribution of the first integral corresponds to the supplementary terms −ρN Nl−1,q−1 , −ρN Nlr−1,q−1 , r = 0, 1 and −ρM Ml−1,q−1 , −ρM Mlr−1,q−1 , r = 0, 1 in cell problems (2.4.7) and boundary layer problems.

102


We obtain thus for all t ∈ (0, T ), the same representation for It (φ) as in Remark 2.2.4, i.e.

It (φ) =

Ψ0,µ (x, t)φ dx +

Cµ

s

m=1

Ψm,µ (x, t)

Cµ

∂φ dx, ∂xm

(2.4.20)

where functions Ψ0,µ , Ψm,µ , m = 1, ..., s, are estimated as sup Ψ0,µ (x, t)L2 (Cµ ) , sup Ψm,µ (x, t)L2 (Cµ ) ≤ cµK

t∈(0,T )

t∈(0,T )

mes Cµ (2.4.21)

with a constant c > 0 independent of µ. Taking φ = u − u(K) , and integrating (2.4.20) from 0 to t in time we obtain the identity

t

It (

0

1 2

=

+

t 0

t

s

Cµ m,j=1

=

Amj

Cµ

s

m=1

x ∂(u − u(K) ) ∂(u − u(K) )

µ

t 0

+

2 ∂ x u − u(K) dxdt + ρ µ ∂t

Cµ

0

Cµ

∂(u − u(K) ) )dt = ∂t

∂xj

∂xm

dxdt, =

Ψ0,µ (x, t)(u − u(K) ) dx +

Ψm,µ (x, t)

∂(u − u(K) ) dx dt. ∂xm

(2.4.22)

It is possible to show by induction that ωj (x1 , t) = 0 for t ≤ θ0 , and therefore

u(K) t=0 = 0.

Taking into account estimate (2.5.21) we estimate the right hand side of identity (2.5.22) by , T K T mes Cµ u − u(K) 2H 1 (Cµ ) dt. c(s + 1)µ 0

Therefore for all t ∈ (0, T ), the integral 2 x 1 u − u(K) dx + ρ µ 2 Cµ

2.5.

103

NON STEADY-STATE ELASTICITY OF A ROD t

+

0

s

Amj

Cµ m,j=1

x ∂(u − u(K) ) ∂(u − u(K) )

∂xm

∂xj

µ

dxdt

is estimated by ,

K

c(s + 1)µ

T mes Cµ

T

u − u(K) 2H 1 (Cµ ) dt.

0

Passing to supt∈(0,T ) in this last estimate we obtain the following inequality κ sup u − u(K) 2L2 (Cµ ) ≤ 2 t∈(0,T ) , T K ≤ c(s + 1)µ T mes Cµ u − u(K) 2H 1 (Cµ ) dt; 0

taking t = T, we have

T

0

s

Amj

Cµ m,j=1

K

≤ c(s + 1)µ

x ∂(u − u(K) ) ∂(u − u(K) )

,

µ

∂xm

∂xj

T mes Cµ

0

dxdt ≤

T

u − u(K) 2H 1 (Cµ ) dt.

So, u − u(K) H 1,0 (Cµ ×(0,T )) =

Theorem 2.4.1 is proved.

= O(µK ) mes Cµ .

2.5

Non steady-state elasticity of a rod

2.5.1


Consider a non steady-state elasticity theory system of equations −ρ(x/µ)

3 ∂u ∂ ∂2u = Fµ (x, t), Amj (x/µ) + 2 ∂xj ∂xm ∂t m,j=1

x ∈ Cµ ,

(2.5.1)


x ∈ ∂U Uµ ,

(2.5.2)

104

CHAPTER 2. HETEROGENEOUS ROD u = 0,

x1 = 0 or x1 = b.

(2.5.3)

and with junction conditions [u] = 0;

s

i,j=1


(2.5.4)

(as in section 2.3) and with initial conditions

u t=0 = 0,

∂u

= 0.

∂t t=0

(2.5.5)

Here Aij (ξ) are matrix functions from section 2.3 and the function ρ(ξ) is 1periodic with respect to ξ1 , ρ(ξ) ≥ κ > 0, and ρ(ξ) is piecewise smooth in the sense of section 2.3. Here κ is the same as in section 3. Besides, ρ(SA ξ) = ρ(ξ) for A = 2, 3 and ′ x x ψ(x1 , t), F Fµ (x, t) = Φ µ µ

where Φ and F are the same as in section 2.3 (F is diagonal), and the fourdimensional vector ψµ is ψµ = (ψ 1 (x1 , t), µ2 ψ 2 (x1 , µt), µ2 ψ 3 (x1 , µt), ψ 4 (x1 , t))∗ , ψ k (x1 , θ) ∈ C ∞ ,

ψ k (x1 , θ) = 0 for θ ≤ θ0 ,

θ0 > 0, k = 1, 2, 3, 4.

It means that the right hand side has two different time-scalings: for the longitudinal and for the torsional components time scale is of order of unity while for two bending components a change θ = µt is done.

2.5.2

Inner expansion

An inner f.a.s. of problem (2.5.1)-(2.5.5) is sought as a series u(∞) =

∞

µl+q Nlq (x/µ)

l+q=0

+

∞

l+q=0

µl+q+2 Mlq (x/µ)

∂ l+q ω(x1 , t) + ∂xl1 ∂tq

∂ l+q ψµ (x1 , t) , ∂xl1 ∂tq

(2.5.6)

where ω is a four-dimensional vector of the form (ω 1 (x1 , t), ω 2 (x1 , µt), ω 3 (x1 , µt), ω 4 (x1 , t))T , Nlq and Mlq are analogs of the functions Nl and Ml from section 2; Nlq are (1-periodic with respect to ξ1 ) solutions of the problems Lξξ Nlq = −T TlqN (ξ) + ΦhN lq ,

ξ ′ ∈ β,

(2.5.7)

2.5.

105

NON STEADY-STATE ELASTICITY OF A ROD 3 ∂N Nlq =− Am1 Nl−1,q nm , ∂ννξ m=2

[N Nlq ] Σ = 0,

TlqN (ξ) =

-

ξ ′ ∈ ∂β,

(2.5.8)

.

3

∂N Nlq

= − A N n

, m1 l −1,q m ∂ννξ Σ Σ m=1

(2.5.9)

3 3 ∂N Nl−1,q ∂ (Aj1 Nl−1,q )+ A1j +A11 Nl−2,q −ρN Nl,q−2 , (2.5.10) ∂ξξj ∂ξξj j=1 j=1

hN lq

=

Φ

∗

3 j=1

∂N Nl−1,q + A11 Nl−2,q − ρN Nl,q−2 A1j ∂ξξj

,

(2.5.11)

Mlq are solutions of the same problems with Nlq replaced by Mlq . However, M00 is a solution of problem (2.3.15). As in section 2.3, we obtain a homogenized equation ∞

µl+q−2 hN lq

l+q=2

∞ l+q ∂ l+q ω ψµ l+q M ∂ − F¯ ψµ = 0, + µ h lq l ∂xl1 ∂tq ∂x1 ∂tq l+q=0

(2.5.13)

whose leading term has the form −ρ

µ2 2

µ

−

2 1 ∂ 2 ω1 ¯ (3) ∂ ω − F¯11 ψ 1 (x1 , t) + · · · = 0, +E 2 ∂x21 ∂t

− ρ

4 2 ∂ 2 ω2 ¯(3) ∂ ω − F¯22 ψ 2 (x1 , θ) + · · · − J 2 ∂θ2 ∂x41

4 3 ∂ 2 ω3 ¯(3) ∂ ω − F¯33 ψ 3 (x1 , θ) + · · · − ρ − J 3 ∂x41 ∂θ2

= 0,

= 0,

2 4 ρ(ξ22 + ξ32 ) ∂ 2 ω 4 (3) ∂ ω − F 44 ψ 4 (x1 , t) + · · · = 0, M + 2 2 ∂x21 ξ2 + ξ3 ∂t2

(2.5.14)

where θ = µt. Let us seek f.a.s. ω of (2.5.13) in the form of a regular series in powers of µ: ∞ (2.5.15) ω= µj ωj . j=0

Substituting (2.5.15) into (2.5.13) we obtain as in section 2.3 a chain of hyperbolic equations −ρ

∂ 2 ωj1 ∂ 2 ωj1 ¯ (3) = fj1 (x1 , t), +E 2 ∂x21 ∂t

−ρ

4 2 ∂ 2 ωj2 (3) ∂ ωj = fj2 (x1 , θ), − J¯2 2 ∂x41 ∂θ

106

CHAPTER 2. HETEROGENEOUS ROD −ρ

−

∂ 4 ωj3 ∂ 2 ωj3 ¯(3) = fj3 (x1 , θ), − J 3 ∂x41 ∂θ2

2 4 ρ(ξ22 + ξ32 ) ∂ 2 ωj4 (3) ∂ ωj M = fj4 (x1 , t), + 2 2 ∂x21 ξ2 + ξ3 ∂t2

(2.5.16)

where right hand sides fjl are some linear combinations of derivatives of functions ωrl with r < j and f0l = F¯ll ψ l .

2.5.3


The boundary layer corrector can be constructed similarly to that in section 2.3. The f.a.s. to problem (2.5.1)–(2.5.5) is sought in the form, u(∞) = uB + u0P + u1P ,

(2.5.17)


∞

0 µl+q Nlq (x/µ)

l+q=0

∞

+


l+q=0

u1P =

∞

+


l+q=0

∂ l+q ψµ (x1 , t)

,


1 µl+q Nlq (x/µ)

l+q=0

∞

∂ l+q ω(x1 , t) + ∂xl1 ∂tq

∂ l+q ω(x1 , t) + ∂xl1 ∂tq

∂ l+q ψµ (x1 , t) |ξ1 =(x1 −b)/µξ′ =x′ /µ ., ∂xl1 ∂tq

(2.5.18)

r r and Nlq , Mlq are exponentially decaying in ξ1 , 1-periodic in all variables except of ξ1 functions. They are found from the chain of boundary value problems similar to that of section 3, but right hand sides of the boundary layer problems contain r r the supplementary term −ρN Nl,q Ml,q −2 , r = 0, 1 (or respectively −ρM −2 , r = 0, 1).

2.5.4

Justification

Theorem 2.5.1. Let K ∈ {0, 1, 2, ...}. Let χ be a function from section 2.3. Consider a function (K)

0(K)

u(K) (x, t) = uB (x, t) + uP

1(K)

(x, t)χ(x1 ) + uP

(x, t)χ(b − x1 ) + ˜(x1 , t),

where (K)

uB

=

K+1

l+q=0

µl+q Nlq

x ∂ l+q ω(x , t) 1 + µ ∂xl1 ∂tq

2.5.

107

NON STEADY-STATE ELASTICITY OF A ROD

K−1

+

µl+q+2 Mlq

l+q=0

u0P =

K+1

x ∂ l+q ψ (x , t) µ 1 , µ ∂xl1 ∂tq

0 µl+q Nlq (x/µ)

l+q=0

+

K−1


l+q=0

u1P =

K+1

∂ l+q ψµ (x1 , t)

,


1 µl+q Nlq (x/µ)

l+q=0

+

K−1


l+q=0

∂ l+q ω(x1 , t) + ∂xl1 ∂tq

∂ l+q ω(x1 , t) + ∂xl1 ∂tq

∂ l+q ψµ (x1 , t) |ξ1 =(x1 −b)/µξ′ =x′ /µ ., ∂xl1 ∂tq

ω (K) =

K

µj ωj (x1 );

j=0

ρ˜(x1 , t) = (1 − x1 /b)q0 (t) + (x1 /b)q1 (t),

qr (t) =

2K+1

µl+q−2

l+q=K+1

+µK+1

r hN pα

j+p+α=l,0≤j≤K,0≤p+α≤K+1

r hM lq

l+q=K−1

0 1 0 1 Nlq , Nlq , Nlq , Mlq , Mlq , Mlq , ωj

∂ p+α ωj + ∂xp1 ∂tα

∂ K−1 ψµ |x1 =rb , r = 1, 2, ∂xl1 ∂tq

are defined in subsections 2.5.2, 2.5.3. Then

the estimate holds

u(K) − uH 1 (Cµ ×(0,T )) = O(µK ) mes Cµ .

(2.5.19)

Justification of the asymptotic is also similar to that in section 2.3. Namely, it is well known that applying the improving smoothness technique [85],[86],[54] (K)

2

(K)

) ) belong to H 1 (C Cµ × (0, T )). So as and ∂ (u−u it can be shown that ∂(u−u ∂t2 ∂t in section 2.3 we estimate the discrepancy functional

It (φ) =

x ∂ 2 (u − u(K) ) , φ dx + ρ ∂t2 µ Cµ

108


+

s

Amj

Cµ m,j=1

x ∂(u − u(K) )

∂xj

µ

,

∂φ dx, ∂xm

(2.5.20)

where for all t ∈ (0, T ), φ ∈ H 1 (C Cµ ), φ = 0 for x1 = 0 or b. The second integral is developed in the same way as in section 2.3, while the contribution of the first integral corresponds to the supplementary terms r r −ρN Nl,q−2 , −ρN Nl,q Ml,q−2 , −ρM Ml,q −2 , r = 0, 1 and −ρM −2 , r = 0, 1 in cell problems (2.5.7) and boundary layer problems. We obtain thus for all t ∈ (0, T ), the same representation for It (φ) as in Remark 2.2.4, i.e.

It (φ) =

Cµ

s Ψ0,µ (x, t), φ dx + m=1

Cµ

Ψm,µ (x, t),

∂φ dx, (2.5.21) ∂xm

where vector valued functions Ψ0,µ , Ψm,µ , m = 1, ..., s, are estimated as sup Ψ0,µ (x, t)L2 (Cµ ) , sup Ψm,µ (x, t)L2 (Cµ ) ≤ cµK

t∈(0,T )

t∈(0,T )

mes Cµ (2.5.22)

with a constant c > 0 independent of µ. Moreover, one can easily check that the time derivatives of functions Ψ0,µ , Ψm,µ , m = 1, ..., s, satisfy the same estimate, i.e.

sup

t∈(0,T )

∂Ψm,µ (x, t) ∂Ψ0,µ (x, t) L2 (Cµ ) ≤ c˜µK mes Cµ (2.5.23) L2 (Cµ ) , sup ∂t ∂t t∈(0,T )

with a constant c˜ > 0 independent of µ. (K) ) , and integrating (2.5.21) from 0 to t in time we Taking φ = ∂(u−u ∂t obtain the identity

t

It (

0

=

1 + 2

1 2

t 0

Cµ

∂(u − u(K) ) )dt = ∂t

∂ x ∂(u − u(K) ) 2 dxdt + ρ ∂t µ ∂t

t 0

s x ∂(u − u(K) ) ∂(u − u(K) ) ∂ dxdt = , Amj ∂xm ∂xj µ Cµ m,j=1 ∂t

=

t 0

Cµ

Ψ0,µ (x, t),

∂(u − u(K) ) dx + ∂t

2.5.

NON STEADY-STATE ELASTICITY OF A ROD s

+

m=1

Cµ

Ψm,µ (x, t),

∂ ∂(u − u(K) ) dx dt. ∂xm ∂t

109

(2.5.24)

It is possible to show by induction that

and therefore

ωjk (x1 , t) = 0 for k = 1, 4

and t ≤ θ0 ,

ωjk (x1 , θ) = 0 for k = 2, 3

and θ ≤ θ0 ,

u(K) t=0 = 0

∂u(K)

= 0. ∂t t=0

and

Therefore the right hand side of identity (2.5.24) can be integrated by parts with respect to t and it will be transformed then into t ∂ (Ψ0,µ (x, t)), (u − u(K) ) dx + − 0 Cµ ∂t s ∂ ∂(u − u(K) ) dx dt + (Ψm,µ (x, t)), + ∂xm ∂t m=1 Cµ

+

Cµ

s Ψ0,µ (x, t), φ dx + m=1

Cµ

Ψm,µ (x, t),

∂φ dx. ∂xm

(2.5.25)

Taking into account estimates (2.5.22),(2.5.23) we estimate the right hand side of identity (2.5.24) by (˜T ˜ + c)(s + 1)µK

mes Cµ sup u − u(K) H 1 (Cµ ) . t∈(0,T )

Therefore for all t ∈ (0, T ), the integral 1 2

+

1 2

x ∂(u − u(K) ) 2 dx + ρ ∂t µ Cµ

s

Amj

Cµ m,j=1

x ∂(u − u(K) ) ∂(u − u(K) ) dx , ∂xm ∂xj µ

is estimated by (˜T ˜ + c)(s + 1)µK


Passing to supt∈(0,T ) in this last estimate we obtain the following inequality

∂(u − u(K) ) 2 κ L2 (Cµ ) + sup ∂t 2 t∈(0,T )

110


+

1 2

s

Amj

Cµ m,j=1

x ∂(u − u(K) ) ∂(u − u(K) ) , dx ≤ ∂xj ∂xm µ

≤ (˜T ˜ + c)(s + 1)µK

So, we get

1 + 2

s

Cµ m,j=1


∂(u − u(K) ) 2L2 (Cµ ) + ∂t t∈(0,T ) sup

Amj

x ∂(u − u(K) ) ∂(u − u(K) ) 1/2 dx = , ∂xm ∂xj µ

= O(µK ) mes Cµ .

Cµ × (0, T )) in (2.5.19), therefore (2.5.19) is This norm is stronger than H 1 (C the corollary of this estimate. Theorem 2.5.1 is proved.

2.6

Contrasting coefficients

(Multi-component homogenization) In this section we consider the highly contrasting coefficients in a 2D model for a rod. We will find out that the classical homogenization can be applied here only in some partial case, when the ”contrast” is not too high with respect to µ2 . In the opposite case the multi-component homogenization introduced in [146],[147] should be applied. This example shows that the contrast of the coefficients is an important physical parameter which can correlate with the other geometrical parameters. This scale effect (see also [141]) was used for the invention of a method for measuring of the conductivity of highly heterogeneous stratified plate [164]. Consider a model problem of a rod with contrasting coefficients. Assume that 1 1 Kω (y) = ωK+ if y ∈ ( , ), 4 2

1 1 Kω (y) = K0 if y ∈ (− , ), 4 4

1 1 Kω (y) = ωK− if y ∈ (− , − ), 2 4 where K0 , K+ , K− > 0, ω > 0. Consider the conductivity equation

div(K Kω (

µ x2 )∇uµ,ω ) = ωf (x1 ), x1 ∈ IR, |x2 | < , 2 µ

(2.6.1)

111

2.6. CONTRASTING COEFFICIENTS with Neumann condition Kω (

µ x2 ∂uµ,ω = 0, x2 = ± , ) 2 µ ∂x2

(2.6.2)

and the T −periodicity condition in x1 ; f is assumed to be a smooth T −periodic T function of C ∞ (IIR), such that 0 f (x1 )dx1 = 0. Thus we consider the 2-dimensional rod (or plate) constituted of three layers µ µ 2 G+ µ = {x ∈ IR , x2 ∈ ( , )}, 4 2

G0µ = {x ∈ IR2 , |x2 |
0, such that µ = O(ω −α ), ω −1 = O(µβ ). The asymptotic solution is sought in the form of multi-component homogenization ansatz [146],[147]: ∞ ∞ d2l v± (x1 ) ± + u(∞) µ2l ω −k Nlk (ξ2 ) µ,ω = ( dx2l 1 ± l=0 k=0

+

∞ ∞

µ2l ω −k Mlk (ξ2 )

l=0 k=−1

d2l f (x1 ) )|ξ2 = xµ2 , dx2l 1

(2.6.4)

+ − , Nlk , Mlk , v+ and v− are unknown functions. Ansatz (2.6.4) differs where Nlk from Bakhvalov’s ansatz by the presence of two different functions v+ and v− , responsible for the macroscopic description of the upper layer G+ µ and of the lower layer G− µ respectively. Substituting (2.6.4) into equation (2.6.1), into Neumann condition (2.6.2) and into interface conditions (2.6.3) we obtain ∞ ∞ ± d2 Nlk d2l v± (x1 ) ± + ( µ2l−2 ω −k+1 K+ ( + N (ξ )) 2 l−1,k dξ22 dx2l 1 ± l=0 k=0

+

∞ ∞

l=0 k=−1

µ2l−2 ω −k+1 K+ (

d2l f (x1 ) d2 Mlk )|ξ2 = xµ2 = + Ml−1,k (ξ2 )) 2 dξ2 dx2l 1

113

2.6. CONTRASTING COEFFICIENTS 1 1 = ωf (x1 ), ξ2 ∈ ( , ), 4 2

(2.6.5)

∞ ∞ ± d2 Nlk d2l v± (x1 ) ± + N (ξ )) + ( µ2l−2 ω −k+1 K− ( 2 −1,k l dξ22 dx2l 1 ± l=0 k=0

+

∞ ∞

l=0 k=−1

µ2l−2 ω −k+1 K− (

d2l f (x1 ) d2 Mlk + Ml−1,k (ξ2 )) )|ξ2 = xµ2 = 2 dξ2 dx2l 1

1 1 = ωf (x1 ), ξ2 ∈ (− , − ), 2 4

(

∞ ∞

µ2l−2 ω −k

∞ ∞

K0 (

±

l=0 k=0

+

µ2l−2 ω −k K0 (

l=0 k=−1

(2.6.6)

± d2 Nlk d2l v± (x1 ) + + Nl±−1,k (ξ2 )) 2 dξ2 dx2l 1

d2l f (x1 ) d2 Mlk )|ξ2 = xµ2 = + Ml−1,k (ξ2 )) 2 dξ2 dx2l 1

1 1 = ωf (x1 ), ξ2 ∈ (− , ), 4 4

(2.6.7)

∞ ∞ ± 2l dN Nlk d v± (x1 ) + µ2l−1 ω −k+1 ( dξ2 dx2l 1 ± l=0 k=0

+

∞ ∞

µ2l−1 ω −k+1

l=−1 k=0

1 1 dM Mlk d2l f (x1 ) )|ξ2 = xµ2 = 0, ξ2 = − or , (2.6.8) 2 2 dξ2 dx2l 1

∞ ∞ 2l ± d v± (x1 ) + ( µ2l ω −k [N Nlk ] dx2l 1 ± l=0 k=0

+

∞ ∞

µ2l ω −k [M Mlk ]

l=0 k=−1

∞ ∞

µ2l−1 ω −k+1

+

∞ ∞

K+

±

l=0 k=0

−

1 1 d2l f (x1 ) )|ξ2 = xµ2 = 0, ξ2 = − or , 4 4 dx2l 1

µ2l−1 ω −k

l=0 k=0

∞ ∞

l=0 k=−1

±

K0

± dN Nlk d2l v± (x1 ) 1 − ( + 0) dξ2 4 dx2l 1

± dN Nlk d2l v± (x1 ) 1 + ( − 0) dξ2 4 dx2l 1

µ2l−1 ω −k+1 K+

d2l f (x1 ) dM Mlk 1 − ( + 0) dξ2 4 dx2l 1

(2.6.9)

114


−

∞ ∞

µ2l−1 ω −k K0

l=0 k=−1

d2l f (x1 ) dM Mlk 1 = 0, ( − 0) dξ2 4 dx2l 1

(2.6.10)

and the analogous condition for ξ2 = − 41 . ± and Mlk satisfy the following boundary value problems in Require that Nlk 1 1 + G = {ξ2 ∈ ( 4 , 2 )} :

K+ (

± d2 Nlk 1 1 + Nl±−1,k (ξ2 )) = h±,+ lk , ξ2 ∈ ( , ), 4 2 dξ22

± dN Nlk 1 ( ) = 0, dξ2 2

±

(2.6.11)

(2.6.12)

± dN Nl,k−1 1 dN Nlk 1 ( − 0), ( + 0) = K0 4 dξ2 dξ2 4

(2.6.13)

1 1 d2 Mlk + Ml−1,k (ξ2 )) = h+ lk , ξ2 ∈ ( , ), 4 2 dξ22

(2.6.14)

K+

and K+ (

dM Mlk 1 ( ) = 0, dξ2 2

K+

dM Ml,k−1 1 dM Mlk 1 ( − 0), ( + 0) = K0 4 dξ2 dξ2 4

(2.6.15)

(2.6.16)

analogously, in G− = {ξ2 ∈ (− 12 , − 14 )} :

K− (

± d2 Nlk 1 1 + Nl±−1,k (ξ2 )) = h±,− lk , ξ2 ∈ (− , − ), 2 2 4 dξ2

± dN Nlk 1 (− ) = 0, 2 dξ2

±

(2.6.17)

(2.6.18)

± dN Nl,k−1 1 dN Nlk 1 (− + 0), (− − 0) = K0 4 dξ2 4 dξ2

(2.6.19)

1 1 d2 Mlk + Ml−1,k (ξ2 )) = h+ lk , ξ2 ∈ (− , − )}, dξ22 2 4

(2.6.20)

K−

and K− (

dM Mlk 1 (− ) = 0, dξ2 2 K−

dM Mlk 1 dM Ml,k−1 1 (− + 0), (− − 0) = K0 4 dξ2 dξ2 4

(2.6.21)

(2.6.22)

115

2.6. CONTRASTING COEFFICIENTS and in G0 = {ξ2 ∈ (− 14 , 14 )} :

K0 (

K0 (

± d2 Nlk 1 1 + Nl±−1,k (ξ2 )) = 0, ξ2 ∈ (− , ), 2 dξ2 4 4

(2.6.23)

± 1 ± 1 ( − 0) = Nlk Nlk ( + 0) , 4 4

(2.6.24)

1 1 ± ± (− − 0) , (− + 0) = Nlk Nlk 4 4

(2.6.25)

1 1 d2 Mlk + Ml−1,k (ξ2 )) = δl1 δk,−1 , ξ2 ∈ (− , ), 4 4 dξ22

(2.6.26)

1 1 Mlk ( − 0) = Mlk ( + 0) , 4 4

(2.6.27)

1 1 Mlk (− + 0) = Mlk (− − 0) , 4 4

(2.6.28)

± Here and below Nlk = 0 if l < 0 or k < 0; Mlk = 0 if l < 0 or k < −1. Note that Ml,−1 (ξ2 ) = 0 if |ξ2 | > 14 . and h± It is clear that h±,± lk lk should be taken in such a way that problems (2.6.11)-(2.6.13), (2.6.14)-(2.6.16), (2.6.17)-(2.6.19), (2.6.20)-(2.6.22) have a solution, ie.,

h±,+ = 4(K+ Nl±−1,k + − K0 lk

h+ lk = 4(K+ Ml−1,k + − K0

h±,− = 4(K− Nl±−1,k − + K0 lk

h− lk = 4(K− Ml−1,k − + K0

.+ =

1 2

1 4

± dN Nl,k −1 1 ( − 0)), 4 dξ2

dM Ml,k−1 1 ( − 0)), 4 dξ2 ± dN Nl,k −1

dξ2

1 (− + 0)), 4

dM Ml,k−1 1 (− + 0)), 4 dξ2

.dξ2 , .− =

− 14

− 12

.dξ2 .

+ (ξ2 ) = 1 in To initialize the recurrent chain of problems, let us take N00 + − + 1 0 − G , N00 (ξ2 ) = 2(ξ2 + 4 ) in G , N00 (ξ2 ) = 0 in G , and N00 (ξ2 ) = 0 in G+ , − − 1 1 ) (ξ2 ) = 1 in G− , and M00 (ξ2 ) = 2K (ξ22 − 16 N00 (ξ2 ) = −2(ξ2 − 14 ) in G0 , N00 0 in G0 , M1,−1 (ξ2 ) = 0 in G± ; M1,−1 (ξ2 ) = 0 in G± ; M0,−1 (ξ2 ) = 0 everywhere. So, we get +

−,+ ++ +− −− h±,± 00 = 0, h10 = 4K+ + , h10 = h01 = 0, h10 = 4K− − ,

116


h++ K0 , h−+ K0 , h+− K0 , h−− K0 , 01 = −8K 01 = 8K 01 = 8K 01 = −8K + − + − h± 00 = 0, h10 = h10 = −1; h01 = h01 = 0.

Equations (2.6.5) and (2.6.6) transform into 4ωK+ +

d2 v+ − 8µ−2 K0 (v+ − v− ) + S + = 2ωf (x1 ), dx21

(2.6.29)

4ωK− −

d2 v− + 8µ−2 K0 (v+ − v− ) + S − = 2ωf (x1 ), dx21

(2.6.30)

and

where

S+ =

µ2l−2 ω −k+1

h±+ lk

±

l,k≥0;l+k≥2

+

µ2l−2 ω −k+1 h+ lk

l,k≥0;l+k≥2

d2l v± (x1 ) + dx2l 1

d2l f (x1 ) dx2l 1

and

S− =

µ2l−2 ω −k+1

h±− lk

±

l,k≥0;l+k≥2

+

µ2l−2 ω −k+1 h− lk

l,k≥0;l+k≥2

d2l v± (x1 ) + dx2l 1

d2l f (x1 ) . dx2l 1

and v+ and v− are sought as the series v± (x1 ) =

∞

p1 ,p2 ,p3 =0

µ2p1 ω −p2 (µ2 ω)−p3 v±p1 p2 p3 (x1 )

(2.6.31)

if µ2 ω → +∞ and v± (x1 ) =

∞

p1 ,p2 =0

µ2p1 ω −p2 v±p1 p2 (x1 )

(2.6.32)

if µ2 ω = const = κ. We obtain the recurrent chain of problems for functions v±p1 p2 p3 and v±p1 p2 of the form 4K+ +

d2 v+p1 p2 p3 (x1 ) = fp+1 p2 p3 (x1 ), dx21

117

2.6. CONTRASTING COEFFICIENTS

4K− −

d2 v−p1 p2 p3 (x1 ) = fp−1 p2 p3 (x1 ), dx21

(2.6.33)

and

4K+ +

4K− −

d2 v+p1 p2 (x1 ) − 8κ−1 K0 (v+p1 p2 (x1 ) − v−p1 p2 (x1 )) = fp+1 p2 (x1 ), dx21

d2 v−p1 p2 (x1 ) + 8κ−1 K0 (v+p1 p2 (x1 ) − v−p1 p2 (x1 )) = fp−1 p2 (x1 ), (2.6.34) dx21

± ± respectively, where f000 (x1 ) = 2f (x1 ), f00 (x1 ) = 2f (x1 ), and fp±1 p2 p3 , fp±1 p2 depend on v±q1 q1 q2 (respectively on v±q1 q2 ) with qi ≤ pi ( i = 1, 2 and eventually 3), (q1 , q2 , q3 ) = (p1 , p2 , p3 ) and (q1 , q2 ) = (p1 , p2 ). Justification is made as usual by a truncation of the series at the level l ≤ J, k ≤ J, pi ≤ J (i = 1, 2 and eventually 3). We obtain in a standard way: 2J ∇(uµ,ω − u(J) + ω −J + (µ2 ω)−J ) µ,ω )L2 (0,T )×βµ ) ≤ C(µ

(2.6.35)

in the case µ2 ω → +∞ and 2J + ω −J ) ∇(uµ,ω − u(J) µ,ω )L2 (0,T )×βµ ) ≤ C(µ

(2.6.36)

(J)

in the case µ2 ω = const, where uµ,ω is a truncated sum (2.6.4), (2.6.31), (2.6.32) at l = J + 1, k = J, pi = J. The leading term of the expansion is + N00 (

x2 x2 − x2 ( )v−0 (x1 ) + M1,−1 ( )f (x1 ), )v+0 (x1 ) + N00 µ µ µ

where v+0 , v−0 is a solution of the system 2K+ +

d2 v+0 (x1 ) − 4κ−1 K0 (v+0 (x1 ) − v−0 (x1 )) = f (x1 ), dx21

2K− −

d2 v−0 (x1 ) + 4κ−1 K0 (v+0 (x1 ) − v−0 (x1 )) = f (x1 ), dx21

κ = 0 if µ2 ω → +∞. It means that when µ2 ω → +∞ the macroscopic fields v+0 in G+ µ and v−0 in G− µ are independent and different if K+ = K− . When µ2 ω = const there is some interaction between the macroscopic fields of upper and lower layers, nevertheless they are very different if K+ = K− . Let us consider now the third case when µ2 ω → +0. Then the behavior of the solution is close to the case of finite ω (see section 2.2). We seek an asymptotic solution in a form of Bakhvalov’s ansatz

118


∞ d2l v(x1 ) (ω) )|ξ2 =x2 /µ , u(∞) = ( µ2l Nl (ξ2 ) µ,ω dx2l 1 l=0

(ω)

where Nl

(2.6.36)

are solutions of the cell problems

1 1 dN Nl (ω) d (ω) ) + Kω (ξ2 )N Nl−1 (ξ2 ) = hl , ξ2 ∈ (− , ), (K Kω (ξ2 ) dξ2 2 2 dξ2

(2.6.37)

(ω)

dN Nl 1 (± ) = 0, 2 dξ2

(2.6.38)

At the points ξ2 = ± 14 the interface conditions hold true (ω)

[N Nl

(ω)

dN Nl ] = 0, dξ2

] = 0, [K Kω (ξ2 )

(ω)

hl = K Kω (ξ2 )N Nl−1 (ξ2 ),

(2.6.39) (2.6.40)

here . =

1 2

.dξ2 .

− 12

We define N0 = 1. We get

(ω)

Kω (ξ2 )

dN Nl dξ2

=

ξ2

− 12

(ω)

(hl − Kω (y)N Nl−1 (y))dy

and so, (ω)

dN Nl dξ2

1 Kω (ξ2 )

=

ξ2

− 12

(ω)

(hl − Kω (y)N Nl−1 (y))dy.

By induction we can prove that h0 = 0, hl =

l

ω l1 hll1 , l > 0

(2.6.41)

l1 =−l+1

(ω)

Nl

(ξ2 ) =

l

(ω)

ω l1 Nll1 (ξ2 )

(2.6.42)

l1 =−l

hll1 , Nll1 do not depend on µ. Indeed, for l = 0, N0 = 1, h0 = 0; let (2.6.41), (2.6.42) be true for some l; then

119

2.6. CONTRASTING COEFFICIENTS

(ω)

hl+1 = K Kω (ξ2 )N Nl

(ξ2 )) = K Kω (ξ2 ) l+1

=

l

(ω)

ω l1 Nll1 (ξ2 ) =

l1 =−l

ω l1 hll1 ;

l1 =−l

(ω)

dN Nl+1 1 = Kω (ξ2 ) dξ2

l+1

ξ2

(

− 12

l1 =−l

ω l1 hll1 − Kω (y)

l+1

=

l

(ω)

ω l1 Nll1 (y))dy =

l1 =−l

ω l1 Fll1 (ξ2 ),

l1 =−(l+1)

where Fll1 do not depend on ω. For v we get the homogenized equation ∞

µ2l−2 hl

l=1

d2l v(x1 ) = ωf (x1 ), dx2l 1

(2.6.43)

where h1 = ω(K+ + + K− − ) + K0 /2; i.e., ∞ l

µ2l−2 ω l1 −1 hll1

l=1 l1 =−l

d2l v(x1 ) = f (x1 ), dx2l 1

i.e. ∞ 2l

(µ2 ω)l−1 ω −l2 hl,l−l2

l=1 l2 =0

d2l v(x1 ) = f (x1 ), dx2l 1

(2.6.44)

Then v is sought in a form v=

∞

(µ2 ω)j ω −k vjk (x1 ),

(2.6.45)

j,k=0

and we have for vjk the equations (K+ + + K− − )

d2 vjk (x1 ) = fjk (x1 ), dx21

(2.6.46)

and fjk (x1 ) determined by vj1 k1 such that, j1 ≤ j, k1 ≤ k, (j1 , k1 ) = (j, k); f00 = f. The leading term is v00 that is a solution of the equation (K+ + + K− − )

d2 v00 (x1 ) = f (x1 ). dx21

(2.6.47)

120

CHAPTER 2. HETEROGENEOUS ROD (J)

For the truncated sum uµ,ω (2.6.36), (2.6.45) at l ≤ J + 1, j ≤ J, k ≤ J the following estimate holds true : −J ∇(uµ,ω − u(J) + (µ2 ω)J ) µ,ω )L2 ((0,T )×βµ ) ≤ C(ω

(2.6.48)

where C does not depend on the parameters. So, in this case, when µ2 ω → +0, the solution tends to function v00 independent of µ and ω. Thus if µ2 ω is a small parameter then problem (2.6.1) - (2.6.3) has a classical homogenization limit v00 . If µ2 ω is a great or finite parameter then there is no a unique limit function v independent of µ and ω : the solution tends in the upper layer to function v+ and in the lower layer to function v− . This is a case of the multi-component homogenization.

2.7

EFMODUL: a code for cell problem s

In asymptotic study of partial differential equations auxiliary problems arise so-called cell problems. The solutions of these problems enable us to define the effective properties of heterogeneous media. These cell problems are solved numerically by the finite-difference method implemented in the code EFMODUL. The numerical results are analyzed and the possibility of an application of the extrapolation method is discussed here below. Consider the system of equations s x ∂u x ∂u x ∂2u ∂ = f (x, t). (2.7.1) A − +S R ij ∂xi ε ∂xj ε ∂t ε ∂t2 i,j=1

Here R(ξ), S(ξ) and Aij (ξ) are piecewise-smooth 1-periodic square n×n matricesvalued functions of the variable ξ ∈ Rs , s = 1, 2, 3; u, f are n−dimensional vectors, n = 1, 2, 3; ε is a small parameter. Equation (2.7.1) describes different processes in composite materials: the wave propagation in elastic non-uniform media, heat transfer (R = 0), diffusion, flow in porous media and so on ; if R = S = 0 we have a steady state problem. An asymptotic study of equation (1) as ε → 0 is given in [16] by the homogenization theory. It is proved that in some assumptions on the coefficients the solution to equation (2.7.1) tends to a solution v of the homogenized equation that is, s ∂v ∂ ∂v ∂2v ˆ ˆ ˆ = f (x, t) (2.7.2) Aij − R 2 +S ∂xj ∂t i,j=1 ∂xi ∂t

with corresponding initial and boundary conditions. The constant n×n matrixvalued coefficients of this equation are called the effective (or macroscopic, or homogenized) coefficients. One of the fundamental problems of the homogenization theory is the calculation of these coefficients. In [16] it was shown that ˆ = R(ξ), Sˆ = S(ξ), where . denotes the integral over a periodic cubic cell R {0 < ξi < 1, i = 1, 2, . . . , s}. To determine Aîj we have to solve s, subsidiary

121

2.7. EFMODUL: A CODE FOR CELL PROBLEMS

cell problems for 1-periodic unknown n × n matrix-valued functions Nq (ξ) (i.e. for every q = 1, 2, . . . , s): s ∂(N Nq + ξq I) ∂ (2.7.3) = 0, ξ ∈ Rs , Aij (ξ) ∂ξξj ∂ξi i,j=1

where I is the unit n × n− matrix, and put 0 / s (N + I) N ξ j j . Aîj = Air (ξ) ∂ξr r=1

(2.7.4)

In the case when the coefficients Aij (ξ) are discontinuous, the solution Nq of systems (2.7.3) is understood in the sense of the variational formulation; in the classical formulation on the surfaces of discontinuity the interface conditions ⎡ ⎤ s ∂(N Nq + ξq I) [N Nq ] = 0, ⎣ Aij (ξ) nj (ξ)⎦ = 0, ∂ξξj i,j=1 are specified, where [.] is the function jump across the surface of discontinuity and nj (ξ) are the direction cosines of the normal vector. In the composite s ∂(N Nj +ξj I) are also of materials mechanics the quantities A˜ij (ξ) = r=1 Air (ξ) ∂ξ r interest since in the case of the elasticity problem the local (microscopic) stresses are calculated from the formula l σik (x) = a ˜kl ij (x/ε)ej (v) + O(ǫ), l j l ˜ where a ˜kl ij (ξ) are elements of the matrix Aij (ξ) and ej (v) = (∂v /∂xl +∂v /∂xj )/2 is the mean (macroscopic) strain tensor (see [16] for a more accurate formulation). For every q, cell problem (2.7.3) for the matrix-valued unknown function can be split into n independent systems relative to the columns Nqp of the matrix Nq , p = 1, 2, . . . , n. We shall rewrite (2.7.3) in the form s ∂ Ai (ξ, ∇ξ (N Nqp + ξq ep )) = 0, p = 1, 2, . . . , n, ∂ξ i i=1

(2.7.5)

where ep is the p-th column of the unit matrix, ∇ξ (N Nqp + ξq E p ) is n × s−matrix p p p p constituted of columns ∂(N N +ξ E )/∂ξ , ..., ∂(N N +ξ q 1 q E )/∂ξs and Ai (ξ, ∇ξ N ) = q q s j=1 Aij ∂N/∂ξi is an n-dimensional vector. A difference scheme of the secondorder approximation (for smooth Ai ) is used to solve system (2.7.5) Lh Nqp = 0

(2.7.6)

where Lh Nqp =

s i=1

(i)

(i)

[Ai (ξ+ , (N Nqp + ξq ep )ξ(i) ) − Ai (ξ− , (N Nqp + ξq ep )ξ(i) ) ]/hi , +

−

(2.7.7)

122


(i)

ξ± = ξ ±ei hi /2, ei = (δi1 . . . , δis )∗ (as above), and for any n-dimensional vector ϕ(ξ) we define the n × s−matrix ϕξ(i) in such a way that, its j-th column is ±

equal to [±ϕ(ξ ± ei hi ) ∓ ϕ(ξ)]/hi if j = i, and [ϕ(ξ ± ei hi + ej hj ) + ϕ(ξ + ej hj ) − ϕ(ξ ± ei hi − ej hj ) − ϕ(ξ − ej hj )]/4hj if j = i (here there is no summing over the repeated index), and hi , . . . , hs are the steps with respect to ξi , . . . , ξs . The mesh is chosen to be uniform : ξ = (i1 h1 , . . . , is hs )t , i1 , . . . , is ∈ ZZ, ∗ is the sign of transposition, and ZZ is the set of integers. In order to solve, in turn, the system of algebraic equations (2.7.6), an implicit fix point method was applied with inversion of the difference analogue of the Laplacian (a fast Fourier transform was used, see [173]) : ΛN Nqp(k) = ΛN Nqp(k−1) − τ Lh Nqp(k−1) ,

(2.7.8)

p(k)

where Nq is the k − th order iteration approximation, τ is the iteration parameter and s Λϕ = ϕξ¯i ξi i=1

is the difference analogue of the Laplace operator [173]. The given method can be used as well in the case of the non-linear system of equations R

x ∂2u

ε

∂t2

+S

x ∂u

ε

∂t

−

s x d Ai , ∇x u = f (x, t) dxi ε i=1

(2.7.9)

where Ai (ξ, y) is 1-periodic in ξ and an n-dimensional vector-valued function that is smooth in y, y is an n × s matrix, ∇x u = (∂u/∂x1 , . . . , ∂u/∂xε ), and u is an n-dimensional vector-valued unknown function. An analogue of the cell problems (2.7.3)is the parametric family of systems of equations (y is the matrix parameter) d Ai (ξ, ∇ξ N (ξ, y) + y) = 0, (2.7.10) dξi

the solution N (ξ, y) is found in the class of n-dimensional vector functions that are 1-periodic in ξ. Homogenized equation has the form (see [16]) 2 ˆ ∂ v + Sˆ ∂v − d Aî (∇x v) = f (x, t), R dxi ∂t ∂t2

(2.7.11)

where ˆ = R(ξ), Sˆ = S(ξ), Aî (y) = Ai (ξ, ∇ξ N + y) R The components of the stress tensor are calculated from the formulae σi ≈ Ai (x/ε, ∇ξ N (x/ε, ∇x v) + ∇x v) To solve (2.7.10) the difference scheme Lh N = 0

(2.7.12)

2.7. EFMODUL: A CODE FOR CELL PROBLEMS

123

with Lh N =

s i=1

(i)

(i)

[A1 (ξ+ , Nξ(i) + y) − Ai (ξ− , Nξ(i) + y)]/hi +

−

(2.7.13)

is applied and the fixed point method of type (2.7.8) is used : ΛN (k) = ΛN (k−1) − τ Lh N (k−1) , where Λ is the difference analogue of the Laplace operator. Consider porous (perforated) media with an ε-periodic in all the xi pores and denote by Aε ) the domain occupied by pores. Then equation (2.7.9) is set in x ∈ Rs \Aε and it is supplied by the Neumann boundary conditions on pore surface ∂Aε . Ai (x/ε, ∇x u)ni (x/ε) = 0 for x ∈ ∂Aε . In this case equation (2.7.10) is set out of pores, i.e. for ξ ∈ IRs \A0 , where A0 = {ξ| εξ ∈ Aε }. On the boundary ξ ∈ ∂A0 the following boundary condition is set: s Ai (ξ, ∇ξ N + y)ni (ξ) = 0. (2.7.14) i=1

Problem (2.7.10) and (2.7.14) was solved by exactly the same method. In (i) (i) this case in (2.7.8) the quantity Ai was considered to be zero if ξ+ (or ξ− ) is in the set A0 The homogenized equation in this case has the form (2.7.11), where f must be replaced by f mes(Q\A0 ), and Q is the unit cube. The higher order cell problems can be presented in the form s ∂ Ai (ξ, ∇ξ N + Y (ξ)) = F (ξ) ∂ξ i i=1

(2.7.15)

where N is a 1-periodic unknown function and Y (ξ), F (ξ) are 1-periodic in ξ (see [16]). These problems can also be solved by the method given above. The cell problems for heterogeneous rods and plates considered in Chapters 2, 3 have also the same form (2.7.15). In the case when the symmetry of coefficients holds with respect to some coordinate plane, the cell problem can be reduced to the problem in the halfcell by using the properties of evenness and oddness of the components of the solution (see [16], Chapter 5). In this case the periodic problems could be reduced to boundary-value ones. The above difference schemes were implemented in a form of the computer solver EFMODUL for conductivity and elasticity problems. Let us consider some results computed by this solver [142]. I. Consider the case of one equation, i.e. n = 1. Let 5 1 for ξ ∈ G1 Aij (ξ) = δij K(ξ), K(ξ) = κ for ξ ∈ G2 ,

124


where G2 is a sub-domain of the periodic cell Q = {ξ | ξi ∈ (0, 1), i = 1, 2, . . . , s}, G1 = Q\G2 . 1. Consider the case s = 2, G2 is a disk of radius 0.25 with its center at the ˆ (see [16]). Below, values of K ˆ that depend point (0.5,0.5). Then Aîj = δij K ˆ are, respectively, on κ are computed: if κ = 0, 2, 5, 10, 20 then the values of K 0.66, 1.13, 1.30, 1.39, 1.43. Note that if κ >> 1 then the iteration parameter τ epl . In this case the values of a ˆri (y) are calculated if y11 = η, yml = 0 (if m = 1 or ˆ22 = l = 1) for different η: if η = 0.01, 0.02, 0.03 then a ˆ11 = 0.035, 0.0694, 0.0998, a 3 2 j ˆ2 , a î = 0 if i = j. 0.11, 0.022, 0.036, a ˆ3 = a

2.8

Bibliographical Remark

The homogenization techniques plays an important part in the applied mathematics and mechanics of the twenty’s century because it has created the mathematical theory of the heterogeneous media and in particular, composite materials and porous media. We note the pioneer works on the asymptotic analysis

2.8. BIBLIOGRAPHICAL REMARK

127

of equations with rapidly oscillating coefficients by E. Sanchez-Palencia [177], E. De Giorgi and S. Spagnollo [49], N. Bakhvalov [12]-[14], J. L. Lions, A. Bensoussan, and G. Papanicolaou [22], O. Oleinik [127], V. Berdichevsky [23], I. Babuska [8], F.Murat and L.Tartar [109], [187] and V. Marchenko and E. Khruslov [103]. In particular, in [13] the series permitting the construction of the complete asymptotic expansion appeared for the first time. The form of the series (2.1.2) is close to the asymptotic series of the averaging technique introduced by N. Krylov, N. Bogoljubov, and Yu. Mitropol’sky [29]. Nowadays the homogenization techniques is presented by two branches: one is the construction of asymptotic expansions and the other is the H-convergence (the G-convergence) and its generalization that is the two-scale convergence. The first branch [16,22,128,129] gives more information about the solution (all correctors, boundary layers, high order estimates) but it demands more smoothness of the data. The second approach [3,65,109,187] requires less of smoothness of data but often the results are formulated as some convergence theorems while error estimates are less informative than in the first approach. The boundary layer techniques appeared in the works by L.Prandtl in the beginning of the twenty’s century and were generalized for the partial derivative equations by M.Vishik and L.Lusternik in [193]. The mathematical asymptotic study of a thin homogeneous rod has been developed in [1,53,111,189]. The complete asymptotic expansion of a thin inhomogeneous rod with symmetry condition A was first constructed in [83,84]. The beams with rapidly varying cross-section were studied in [192]. General anisotropic heterogeneous rods were studied by the asymptotic expansion method in [117] and [101] and by the scaling and then application of the Hconvergence techniques in [108]. The contact problem of two heterogeneous bars was considered in [149,150,169].

Chapter 3

Heterogeneous Plate Here we consider the three-dimensional conductivity problem and the threedimensional elasticity problem in a thin domain (plate) having a heterogeneous structure. We assume that the characteristic size of heterogeneities is of the same order as the ”width” of the domain. Assume that it is much less than the ”length” of the plate that is a characteristic longitudinal size of the domain. So the ratio of the ”width” to the ”length” of the plate is the small parameter of the same order as the ratio of the characteristic size of heterogeneity to the ”length” of the plate. We assume that the ”length” of the plate has a finite value. The simplest geometrical model of a plate is defined as an intersection of µ = {(x1 , x2 ) ∈ IR2 , x3 /µ ∈ (−1/2, 1/2)} orthogonal to the axis a thin layer U x3 with the ”thick” three-dimensional layer {x = (x1 , x2 , x3 ) ∈ IR3 ; x1 ∈ (0, b)} orthogonal to the axis x1 . As in Chapter 2, we assume that µ is a small parameter. The microscopical heterogeneity of the plate is simulated by a special dependence of the coefficients of the material on space variable x = (x1 , x2 , x3 ); these coefficients are functions of x/µ.

129

130

CHAPTER 3. HETEROGENEOUS PLATE

Figure 3.0.1. Heterogeneous plate. An asymptotic analysis of the mathematical model is developed as µ tends to zero. To this end we apply the homogenization technique and the boundary layer technique (cf. Chapter 2). In section 3.1 we study the conductivity equation; an expansion of a solution is constructed and justified and the twodimensional homogenized model of the plate is obtained. In section 3.2 the three-dimensional elasticity problem for a plate is studied. The question about existence of an equivalent homogeneous plate is discussed in section 3.3. Nonsteady state elasticity equation is studied in section 3.4.

3.1

Conductivity of a plate

3.1.1


µ be the layer {(x1 , x2 ) ∈ IR2 , x3 /µ ∈ (−1/2, 1/2)}. The Definition 1.1 Let U ′ = U µ ∩ {x1 ∈ (0, b)} will be called the plate. intersection C µ Consider the conductivity equation Pu ≡

3 ∂ x ∂u = ψ(x1 , x2 ), Aij µ ∂xj ∂xi i,j=1

′ , x∈C µ

(3.1.1)

131

3.1. CONDUCTIVITY OF A PLATE with boundary conditions 3 x x 1 ∂u ∂u 2 1 = 0, , ,± A3j ≡± 2 ∂xj µ µ ∂ν j=1

u = 0,

x1 = 0

or x1 = b,

µ , x ∈ ∂U

(3.1.2) (3.1.3)

and with the condition of T -periodicity with respect to x2 . Here b and T are numbers of the order of 1 that are multiples of µ and Aij (ξ) satisfy the conditions of section 2.2 and are 1-periodic with respect to ξ1 and ξ2 . These elements are assumed to be piecewise smooth functions (in the sense of section 2.2; for β we take the square (−1/2, 1/2) × (−1/2, 1/2)), while ψ is a C ∞ function that is T -periodic with respect to x2 .

µ′ . Figure 3.1.1. Plate C Remark 3.1.1. In this chapter for the reason of simplicity we consider the case when the right side hand function does not depend on the rapid variable x/µ. Of course such dependance could be taken into account in the same way as in the previous chapter.

3.1.2

Inner expansion

We seek a formal asymptotic solution in the form of a series u(∞) =

∞ l=0

µl

|i|=l

Ni

x µ

Di ω,

(3.1.4)

where i = (i1 , . . . , il ) is a multi-index, ij ∈ {1, 2}, ω(x1 , x2 ) is a smooth T periodic with respect to x2 three-dimensional vector function, Ni (ξ) are 3×3 matrix functions that are 1-periodic with respect to ξ1 and ξ2 .

132


Substituting series (3.1.4) in (3.1.1), (3.1.2) and (2.2.4) and grouping terms of the same order, we obtain P u(∞) − ψ =

∞

µl−2

l=0

|i|=l

Hi (ξ)Di ω − ψ,

(3.1.5)

∞ ∂u(∞) µl−1 Gi (ξ)Di ω, = ∂ν l=0

(3.1.6)

|i|=l

where Hi (ξ) = Lξξ Ni + Ti (ξ), Ti (ξ) =

3 j=1

3 ∂N Ni2 ...il ∂ + Ai1 i2 Ni3 ...il , (Aji1 Ni2 ...il ) + Ai1 j ∂ξξj ∂ξξj j=1

Gi (ξ) =

3 3

m=1

Amj

j=1

∂N Ni + Ami1 Ni2 ...il nm . ∂ξξj

Suppose that ξ ∈ IR2 ×

Hi (ξ) = hi ,

1 1 ; − , 2 2

Gi (ξ) = 0,

1 ξ3 = ± , 2

where hi are constants. Assume that [N Ni ] = 0 and [Gi ] = 0 on the surfaces of discontinuity. We obtain the following recurrent chain of problems of the form Ti (ξ) + hi , Lξξ Ni = −T 3

A3j (ξ)

j=1

[N Ni ] Σ = 0,

∂N Ni = −A3i1 Ni2 ...il , ∂ξξj

(3.1.7)

1 ξ3 = ± , 2

- .

3 ∂N

Ni

= − n A N

m mi1 i2 ...il , ∂ννξ Σ Σ m=1

(3.1.8)

(3.1.9)

to determine Ni that are 1-periodic functions with respect to ξ1 and ξ2 ; here (n1 , n2 , n3 ) is an external normal vector. The constant matrices hi are chosen from the solvability conditions for problem (3.1.7)-(3.1.9): hi =

3 j=1

here · =

(−1/2,1/2)3

Ai 1 j

∂N Ni2 ...il + Ai1 i2 Ni3 ...il , ∂ξξj

h∅ = 0, dξ and N∅ = 1.

hi1 = 0;

l ≥ 2, (3.1.10)

133

3.1. CONDUCTIVITY OF A PLATE

Let us give the variational formulation of (3.1.7)-(3.1.9). Let Q be the unit 1 1 cube (0, 1)2 × (− 12 , 12 ). Let Hper ξ1 ,ξ2 (Q) be closure in the norm H (Q) of the space of differentiable on IR2 ×[−1/2, 1/2] functions of ξ = (ξ1 , ξ2 , ξ3 ), 1-periodic 1 in ξ1 and ξ2 . Then the variational formulation is: find Ni ∈ Hper ξ1 ,ξ2 (Q), such that 1 ∀Φ ∈ Hper

=

3

Ami1 (ξ)N Ni2 ...il

m=1

ξ1 ,ξ2 (Q),

−

3

m,j=1

Amj (ξ)

∂N Ni ∂Φ = ∂ξξj ∂ξm

3 ∂N Ni2 ...il ∂Φ + Ai1 i2 Ni3 ...il − hi )Φ . − ( Ai1 j (ξ) ∂ξξj ∂ξm j=1

It could be proved as in section 2.2 (Lemma 2.2.1) that this problem has a solution if and only if (3.1.10) holds true. This solution is defined up to an arbitrary additive constant. Thus, the algorithm for constructing the functions Ni is recurrent. We formally assume that Ni = 0 for |i| < 0, N∅ = 1, and Ni (ξ) is a solution of problem (3.1.7)–(3.1.10) for l > 0. The right-hand side contains Nj with multi-indices j whose length is smaller than |i|. Theorem 3.1.1. The constant matrix of homogenized coefficients A¯ = (hi1 i2 )1≤i1 ,i2 ≤2 is positive definite and it is symmetric. Proof. 1. It follows from (3.1.10) that hi1 i2 =

3

Ai1 j

j=1

=

3

3 ∂ ∂N N i2 (N Ni2 + ξi2 ) + Ai 1 i 2 = Ai1 j ∂ξξj ∂ξξj j=1

∂ ∂ (N Ni1 + ξi1 ) . (N Ni2 + ξi2 ) ∂ξm ∂ξξj

Amj

m,j=1

The last relation arises from the integral identity for problem (3.1.7)–(3.1.9) with |i| = 2. The symmetry of the matrix of coefficients A = (Amj )1≤m,j≤3 implies the symmetry of the matrix of homogenized coefficients A¯ = (hi1 i2 )1≤i1 ,i2 ≤2 . Let us prove that A¯ is positive definite. Let (η1 , η2 ) ∈ IR2 . Consider the sum 2

i1 ,i2 =1

hi1 i2 ηi2 ηi1 =

2 2 ∂ ( (N Ni2 + ξi2 )ηi2 )( (N Ni1 + ξi1 )ηi1 ) ≥ ∂ξξj i =1 i =1

3

Amj

≥κ

3 2 ∂ (N Nk + ξk )ηk )2 ≥ ( ∂ξ ξ j j=1

≥κ

3

m,j=1

2

1

k=1

j=1

(

2 ∂ (N Nk + ξk )ηk )2 = ∂ξξj

k=1

134


=κ

2

ηk2 ,

k=1 k ∂N ∂ξj

= 0. because The proof of Theorem 3.1.1 is complete. Then (3.1.5) takes the form P u(∞) − ψ =

+

∞

µl−2

l=3

2

hi 1 i 2

i1 ,i2 =1

hi

|i|=l

∂2ω + ∂xi1 ∂xi2

∂lω − ψ = 0. ∂xi1 · · · ∂xil

(3.1.11)

Problem (3.1.11) can be regarded as the homogenized equation of infinite order with respect to the three-dimensional vector ω. A formal asymptotic solution (f.a.s.) of this problem is sought in the form of a series ω=

∞

µj ωj (x1 , x2 ),

(3.1.12)

j=0

where ωj do not depend on µ. Substituting the series (3.1.12) into (3.1.11), we obtain a recurrent chain of equations for the components ωjk of the vectors ωj in the form 2 ∂ 2 ωj A¯i1 i2 = fj (x1 , x2 ), (3.1.13) ∂xi1 ∂xi2 i ,i =1 1

2

where A¯i1 i2 = hi1 i2 , the functions fj depend on ωj1 , j1 < j, and on the derivatives of these functions. Thus, the f.a.s. of the problem (3.1.1), (3.1.2), and (2.2.4) is constructed. The asymptotic analysis of the conductivity equation for inhomogeneous plates was first carried out by D.Caillerie [31], where the homogenized equation of zero order was obtained. Here below we construct the complete asymptotic expansion of a solution of the conductivity equation (taking into account the boundary layers).

3.1.3


A formal asymptotic solution of problem (3.1.1)–(3.1.3), (2.2.4) will be sought again in the form (2.2.22): u(∞) = uB + u0P + u1P , where uB is defined by (3.1.4) and ∞

u0P = µl Ni0 (ξ)Di ω ξ=x/µ , (3.1.14) l=0

|i|=l

135

3.1. CONDUCTIVITY OF A PLATE u1P =

∞

µl

l=0

|i|=l

Ni1 (ξ)Di ω ξ=(x

1 −b)/µ,

ξ ′ =x′ /µ

.

(3.1.15)

Here the functions Ni0 (ξ) and Ni1 (ξ) are 1-periodic with respect to ξ2 and exponentially stabilize to zero as |ξ1 | → +∞. Substituting (3.1.14) and (3.1.15) into (3.1.1)–(3.1.3) and taking into account the fact that uB is constructed above, we obtain recurrent chains of problems to determine Ni0 (ξ) as in subsection 3.1.2 1 1 0 0 ′ , Lξξ Ni + Ti (ξ) = 0, ξ1 > 0, ξ ∈ IR × − , 2 2 3 j=1

A3j

∂N Ni0 = −A3i1 Ni02 ...il , ∂ξξj

1 ξ3 = ± , 2

Ni (0, ξ ′ ) + h0i Ni0 (0, ξ ′ ) = −N

(3.1.16)

Ni1 (ξ)

and obtain similar problems for defined for ξ1 < 0 with the constant h1i instead of h0i . hri are constants chosen from the condition that solutions of problems (3.1.16) stabilize to zero as ξ1 → +∞ (ξ1 → −∞ for r = 1), and N∅0 = 0. Theorem 3.1.2. Let Aij (ξ) satisfy the conditions formulated at the beginning of this section, let the three-dimensional vectors Fj (ξ) be 1-periodic with respect to ξ2 and belong to L2 ([0, +∞) × β), j = 0, 1, 2, 3, let β = (−1/2, 1/2)2 , and let u0 (ξ ′ ) be 1-periodic with respect to ξ2 and belong to H 1/2 ({0} × (−1, 1) × (−1/2, 1/2)). Then a solution u(ξ) of the problem 3 1 1 ∂F Fj ′ , , ξ1 > 0, ξ ∈ IR × − , Lξξ u = F0 (ξ) + ∂ξξj 2 2 j=1

3 j=1

A3j

∂u = Fj , ∂ξξj

1 ξ3 = ± , 2

u ξ1 =0 = u0 (ξ ′ )

(3.1.17)

such that u is 1-periodic with respect to ξ2 and inequality (2.2.36) holds for constant h, exists and is unique. Here by a solution of problem (3.1.17) we mean a function u(ξ) satisfying the integral identity from subsection 3.1.2. Theorem 3.1.2 can be proved similarly to Theorems 4 and 5 in [130] and theorems of the Phragmen-Lindelof ¨ type [89],[90]. The constants hri are defined r similarly to the constants hN from Chapter 2. In this case hr∅ = 1 since N∅r = 0 l and N∅ = 1. After determining all Nir and hri by induction on |i|, we obtain a homogenized problem for the three-dimensional vector function ω(x1 , x2 ): this problem is formed by equation (3.1.14) with the boundary conditions

∞

ω+ µl hri Di ω

= 0. l=1

|i|=l

x1 =rb

136


A f.a.s. of problem (3.1.1)–(3.1.3) is sought in the form of series (3.1.12) with boundary conditions

ωj x =rb = gjr , r = 0, 1. (3.1.18) 1

where the gjr depend on ωjr1 with j1 < j and on derivatives of these functions. The homogenized problem of zero order is formed by relations (3.1.13), (3.1.18) with j = 0, f0 (x1 , x2 ) = ψ(x1 , x2 ),

(3.1.19)

g0r = 0.

The estimate for the difference of the exact solution and the asymptotic solution can be obtained as in Chapter 2.

3.1.4

Algorithm for calculating the effective conductivity of a plate

As in section 2.3, we present a calculation algorithm for the coefficients A¯i1 i2 = hi1 i2 , i1 , i2 ∈ {1, 2}, of the homogenized equation (the effective conductivity tensor in a plane). We first solve the conductivity equations for i1 = 1, 2:

∂ ∂ (N Ni1 + ξi1 ) = 0, Amj ∂ξξj ∂ξm m,j=1 3

3 j=1

A3j

∂ (N Ni1 + ξi1 ) = 0, ∂ξξj

ξ3 ∈

1 1 , − , 2 2

1 ξ3 = ± , 2

(3.1.20)

where Ni1 (ξ) is an unknown function that is 1-periodic with respect to ξ1 and ξ2 . The effective conductivity coefficients in a plane can be defined by the following formulas: A¯i1 i2 = hi1 i2 =

3 j=1

Ai1 j

∂ (N Ni2 + ξi2 ) . ∂ξξj

(3.1.21)

For constant coefficients in the isotropic case Amj = δmj K, K = const, we have: Nk = 0, k = 1, 2, A¯i1 i2 = δi1 i2 K; for constant coefficients in the general case we have A¯i1 i2 = Ai1 i2 − Ai1 3 A−1 33 A3i2 . For coefficients depending on ξ3 only we have

(3.1.22)


137

A¯i1 i2 = Ai1 i2 − Ai1 3 A−1 33 A3i2 ,

1/2 where · = −1/2 dξ3 . Indeed, in this case a solution of problem (3.1.20) is sought as a function of ξ3 only. We have: A33

∂N N i2 + A3i2 = 0, ∂ξ3

so ∂N N i2 = −A−1 33 A3i2 , ∂ξ3

and so (3.1.21) implies

∂N N i2 + Ai1 i2 = Ai1 i2 − Ai1 3 A−1 A¯i1 i2 = Ai1 3 33 A3i2 . ∂x3

Remark 3.1.2. In mechanics it is usual to take µA¯ij as the effective conductivity coefficients in a plane, where µ is the thickness of a plate. The homogenized problem of zero order has the form ∂ ¯ ∂ω0 = f (x1 , x2 ), Amj ∂xj ∂xm m,j=1 2

ω0 = 0 for x1 = 0, b, i.e. ∂ ¯ ∂ω0 = µf (x1 , x2 ), µAmj ∂xj ∂xm m,j=1 2

ω0 = 0 for x1 = 0, b. Evidently, for any given positive definite symmetric matrix µA¯ij , i, j = 1, 2 it is possible to find such a constant positive definite symmetric matrix Aij , i, j = 1, 2, 3 that it is related with the given matrix by equality (3.1.22), for example, Aij = A¯ij , i, j = 1, 2, A33 > 0, and Ai3 , A3j = 0, i, j = 1, 2, 3. It means that any heterogeneous conductive plate (3.1.1)–(3.1.3) is ”equivalent” in the sense of the effective conductivity coefficients in a plane to some homogeneous plate. It is remarkable that the analogous property for elasticity coefficients is not true. This question will be discussed in the section 3.3.

138


3.1.5

Justification of the asymptotic expansion

Here we prove the following main theorem. Theorem 3.1.3. Let K ∈ {0, 1, 2, ...}. Denote χ a function from C (K+2) ([0, b]) such that χ(t) = 1 when t ∈ [0, b/3], χ(t) = 0 when t ∈ [2b/3, b]. Consider a function (K)

0(K)

u(K) (x) = uB (x) + uP

1(K)

(x)χ(x1 ) + uP

(x)χ(b − x1 ) + ρ(x1 , x2 ),

where (K)

uB

=

K+1

µl

l=0

0(K)

uP

=

K+1

µl

l=0

1(K)

uP

=

K+1 l=0

|i|=l

Ni

|i|=l

|i|=l

µl

x

µ

Di ω (K) (x1 , x2 ),

Ni0 (ξ)Di ω (K) (x1 , x2 )

ξ=x/µ

,

Ni1 (ξ)Di ω (K) (x1 , x2 ) |ξ1 =(x1 −b)/µξ′ =x′ /µ ,

ω (K) =

K

µj ωj (x1 , x2 );

j=0

ρ(x1 , x2 ) = (1 − x1 /b)q0 (x2 , µ) + (x1 /b)q1 (x2 , µ),

qr (x2 , µ) =

2K+1

µl

l=K+1

j+p=l,0≤j≤K,0≤p≤K+1 |i|=p

r i hN i D ωj |x1 =rb , r = 1, 2,

Nl , Nl0 , Nl1 , ωj are defined in subsection 2.3. Then the estimate holds u(K) − uH 1 (C˜µ ) = O(µK ) mes C˜µ .

Proof. Let us estimate the discrepancy functional I(φ) =

s

˜µ C m,j=1

Amj

x ∂(u − u(K) ) ∂φ dx, ∂xm ∂xj µ

where φ ∈ H 1 (C Cµ ), φ = 0 for x1 = 0 or b. 1. At the first stage represent it in the form

139


I(φ) =

˜µ C

F ψφ dx − JB (φ) − J0 (φ) − J1 (φ) − J2 (φ),

where

JB (φ) =

J0 (φ) =

J1 (φ) =

s

˜µ C m,j=1

s

˜µ C m,j=1

s

˜µ C m,j=1

=

Amj

J2 (φ) =

Amj

˜µ C

so,

x ∂u(K) ∂φ B dx, µ ∂xj ∂xm

x ∂(u0(K) (x)χ(x )) ∂φ 1 P dx, ∂xj ∂xm µ

x ∂(u1(K) (x)χ(b − x )) ∂φ 1 P dx, ∂xm ∂xj µ

s

˜µ C m,j=1

Amj

Amj

x ∂ρ(x ) ∂φ 1 dx = µ ∂xj ∂xm

s x ∂φ q1 − q0 dx; Am1 µ ∂xm b m=1

|J J2 (φ)| ≤ c0 µK+1 φH 1 (C˜µ )

mes C˜µ .

The boundary layer functions Nlr decay exponentially as |ξ1 | → +∞, so (2.2.36) implies the following estimates

|

|

Amj

s

Amj

˜µ ∩{x1 ≥b/3} C m,j=1

x ∂(u0(K) (x)χ(x )) ∂φ 1 P dx| ≤ c1 e−c2 /µ φH 1 (C˜µ ) , ∂xm ∂xj µ

s

˜µ ∩{x1 ≤2b/3} C m,j=1

x ∂(u1(K) (x)χ(b − x )) ∂φ 1 P dx| ≤ c1 e−c2 /µ φH 1 (C˜µ ) ∂xm ∂xj µ

with some constants c1 , c2 > 0 independent of µ. So I(φ) = F ψφ dx − JB (φ) − Jˆ0 (φ) − Jˆ1 (φ) + ˜µ C

+ (Jˆ0 (φ) − J0 (φ)) + (Jˆ1 (φ) − J1 (φ)) − J2 (φ), where Jˆr (φ) =

s

˜µ C m,j=1

Amj

x ∂ur(K) (x) ∂φ P dx, r = 1, 2, ∂xm ∂xj µ

140


and |Jˆr (φ) − Jr (φ)| ≤ c1 e−c2 /µ φH 1 (C˜µ ) . Consider now JB (φ) = =

K+1

s

˜µ C

l=0 m=1

+Ami1

+

s

m=1

x

µ

Ni2 ...il

Amj

j=1

|i|=l

µK+1

˜µ C

s

µl−1

x µ

Ami1

|i|=K+2

x ∂N Ni (ξ) |ξ=x/µ + ∂ξξj µ

Di ω (K)

x µ

∂φ dx + ∂xm

Ni2 ...iK+2 Di ω (K)

∂φ dx. ∂xm

Here i = (i1 ...il ), ij ∈ {1, 2}. Denote s

ABN mi (ξ) =

Amj (ξ)

j=1

∆N 1 (φ) =

s

m=1

˜µ C

µK+1

∂N Ni (ξ) + Ami1 (ξ)N Ni2 ...il (ξ), ∂ξξj

Ami1

|i|=K+2

x µ

Ni2 ...iK+2 Di ω (K)

∂φ dx. ∂xm

If necessary the subscript m at the notation ABN mi can be ”included” in the BN multi-index i, i.e., if i = (i1 , ..., il ) then ABN , mi can be written in the form A˜i ˜ ˜ where i = (m, i1 , ..., il ); in this case |i| = l + 1. Of course this ”inclusion rule” is true if and only if m ∈ {1, 2}. The boundary layer functions Nir2 ...iK+2 decay exponentially as |ξ1 | → +∞, so K+1 |∆N φH 1 (C˜µ ) mes C˜µ , 1 (φ)| ≤ c3 µ

with a constant c3 > 0 independent of µ. We get

JB (φ) = =

K+1

s

l=0 m=1

˜µ C

µl−1

i (K) ABN mi (x/µ)D ω

|i|=l

+ ∆N 1 (φ). 2. At the second stage consider the sum

∂φ dx + ∂xm

141


N J˜B (φ) = K+1

=

s

˜µ C

l=0 m=1

=

s

K+1

˜µ C

l=0 m=1

−

i (K) ABN (x1 , x2 ) mi (x/µ)D ω

|i|=l

µl−1


|i|=l

K+1

˜µ C

l=0

µl−1

µl−1 φ

2

ABN mi (x/µ)

m=1 |i|=l

∂φ dx = ∂xm

∂ i (K) φD ω dx − ∂xm

∂ Di ω (K) dx. ∂xm

Taking into account the mentioned above ”inclusion rule” for subscripts we 2 ∂ i can replace summation m=1 |i|=l ABN mi ∂xm D by the equivalent summation BN i D and therefore |i|=l+1 Ai N (φ) = J˜B

=

K+1

s

˜µ C

l=0 m=1

−

=

K+1

−

l=0

˜µ C

˜µ µ−1 C


µl−1

˜µ µ−1 C

∂ i (K) φD ω dx − ∂xm

ABN (x/µ)φDi ω (K) dx = i

|i|=l+1

µs+l−2

K+1 l=0

|i|=l

K+1

s

l=0 m=1

µl−1

ABN mi (ξ)

|i|=l

µs+l−1

|i|=l+1

∂ i (K) φD ω |x=µξ dξ − ∂ξm

ABN (ξ) φDi ω (K) |x=µξ dξ, i

where 1 1 µ−1 C˜µ = {ξ ∈ IRs | µξ ∈ C˜µ } = (0, b/µ) × (0, T /µ) × (− , ). 2 2

Finally

N J˜B (φ) =

K+1 l=0

˜µ µ−1 C

µs+l−2

s

|i|=l m=1

ABN mi (ξ)

∂ (φDi ω (K) |x=µξ − ∂ξm

142

CHAPTER 3. HETEROGENEOUS PLATE − ABN (ξ) φDi ω (K) |x=µξ dξ − i

− µs+K

˜µ µ−1 C |i|=K+2

ABN (ξ) φDi ω (K) |x=µξ dξ, i

where ABN mi = 0 when |i| < 0 (by convention). Denote s+K BN i (K) ∆N (φ) = µ A (ξ) φD ω |x=µξ dξ. 2 i ˜µ µ−1 C |i|=K+2

3. The variational formulation for problem (3.1.7)-(3.1.9) gives: 1 ˜ ∀φ(ξ) ∈ Hper

s

m=1

ABN mi (ξ)

ξ1 ξ2 (Q),

∂ φ˜ ˜ ˜ − ABN (ξ) φ(ξ) = hN i i φ(ξ). ∂ξm

1 Then this identity holds, in particular, for any function φ˜ ∈ Hper ξ1 ξ2 (Q), 2 vanishing when (ξ1 , ξ2 ) ∈ ∂SQ , whereSQ is a unit square (0, 1) . So, this identity holds for any function φ˜ ∈ H 1 (Q), vanishing when (ξ1 , ξ2 ) ∈ ∂SQ , as well as for any function φ˜ ∈ H 1 ((a1 , a1 + 1) × (a2 , a2 + 1) × (− 12 , 12 )), a1 , a2 ∈ IR vanishing when (ξ1 , ξ2 ) ∈ ∂SQa1 ,a2 , whereSQa1 ,a2 is a square (a1 , a1 + 1) × (a2 , a2 + 1). In the last case the average is defined as (a1 ,a1 +1)×(a2 ,a2 +1)×(− 1 , 1 ) dξ. 2 2 Let φ˜ be a function from H 1 (µ−1 C˜µ ) vanishing when ξ1 = 0 and

per ξ2 =T /µ

1 −1 ˜ when ξ1 = b/µ. ( Here Hper Cµ ) is the completion with respect to ξ2 =T /µ (µ 1 −1 ˜ the norm H (µ Cµ ∩ {0 < ξ2 < T /µ}) of the set of infinitely differentiable on the closed layer µ−1 C˜µ , T /µ−periodic in ξ2 functions. ) We represent it in the form

˜ ˜ ˜ cos2 (πξ1 )sin2 (πξ2 ) + φ(ξ) = φ(ξ) sin2 (πξ1 )sin2 (πξ2 ) + φ(ξ) ˜ ˜ + φ(ξ) sin2 (πξ1 )cos2 (πξ2 ) + φ(ξ) cos2 (πξ1 )cos2 (πξ2 ), where ˜ sin2 (πξ1 )sin2 (πξ2 ) vanishes when (ξ1 , ξ2 ) ∈ ∂SQ φ(ξ) , a1 ∈ {0, 1, ..., b/µ}, a1 ,a2 a2 ∈ {0, 1, ..., T /µ, } ˜ cos2 (πξ1 )sin2 (πξ2 )vanishes when (ξ1 , ξ2 ) ∈ ∂SQ φ(ξ) , a1 ∈ {0, 1/2, 3/2, ..., a1 ,a2 b/µ−1/2, b/µ}, a2 ∈ {0, 1, ..., T /µ, } ˜ sin2 (πξ1 )cos2 (πξ2 ) vanishes when (ξ1 , ξ2 ) ∈ ∂SQ φ(ξ) , a1 ∈ {0, 1, ..., b/µ}, a1 ,a2 a2 ∈ {−1/2, 1/2, 3/2, ..., T /µ − 1/2, T /µ + 1/2}, } ˜ cos2 (πξ1 )cos2 (πξ2 ) vanishes when (ξ1 , ξ2 ) ∈ ∂SQ φ(ξ) , a1 ∈{0, 1/2, 3/2, ..., a1 ,a2 b/µ−1/2, b/µ}, a2 ∈ {−1/2, 1/2, 3/2, ..., T /µ − 1/2, T /µ + 1/2}.}

143


Therefore, by simple addition the integral identity of the variational formulation for problem (3.1.7)-(3.1.9) can be generalized as

s

˜µ µ−1 C m=1

ABN mi (ξ)

∂ φ˜ ˜ dξ = − ABN (ξ)φ(ξ) i ∂ξm

˜µ µ−1 C

˜ hN i φ(ξ) dξ

1 ˜ (µ−1 C˜µ ), vanishing when ξ1 = 0 or ξ1 = b/µ; for for any φ(ξ) ∈ Hperξ 2 =T /µ ˜ instance, it remains valid for φ(ξ) = φDi ω (K) (x1 , x2 ) |x=µξ .

Thus,

N J˜B (φ) =

K+1

˜µ µ−1 C |i|=l

l=0

=

K+1 l=0

˜µ C |i|=l

i (K) φD µs+l−2 hN ω |x=µξ dξ + ∆N i 2 (φ) =

i (K) µl−2 hN dx + ∆N l φ(x)D ω 2 (φ),

where

(N )

K |∆N ˜µ ) 2 (φ)| ≤ c4 µ φH 1 (C

(N )

mes Cµ ,

with a constant c4 > 0 independent of µ. 0N 4. Replacing Ni by Ni0 in the expressions ABN mi we define Ami . Applying the same reasoning as at stages 2 and 3 we obtain the relation

Jˆ0 (φ) =

K+1

˜µ µ−1 C |i|=l

l=0

− µs+K

µs+l−2

s

A0N mi (ξ)

m=1

∂ (φDi ω (K) |x=µξ − ∂ξm

i (K) − A0N |x=µξ dξ − i (ξ) (φD ω

dK+2 ω (K) (x ) 1 A0N |x=µξ dξ. i (ξ) φ K+2 ˜µ dx µ−1 C 1 |i|=K+2

Variational formulation for problem (3.1.16) gives

|i|=l

s

˜µ µ−1 C m=1

A0N mi (ξ)

∂ φ˜ ˜ − A0N i (ξ)φ(ξ) dξ = 0 ∂ξm

1 for any function φ˜ ∈ Hper(ξ (µ−1 C˜µ ) vanishing when ξ1 = 0 or ξ1 = b/µ. 2 ,T /µ) So, |Jˆ0 (φ)| ≤ c5 µK φ 1 ˜ mes C˜µ , H ( Cµ )

with a constant c5 > 0 independent of µ. Similarly, |Jˆ1 (φ)| ≤ c6 µK φ

˜µ ) H 1 (C

mes C˜µ ,

144


with a constant c6 > 0 independent of µ. Thus,

I(φ) =

˜µ C

(ψ −

K+1

µl−2

l=0

i (K) hN )φ(x) dx + ∆3 (φ), i D ω

|i|=l

where ∆3 (φ) is a linear functional of φ such that |∆3 (φ)| ≤ c7 µK φH 1 (C˜µ ) mes C˜µ ,

with a constant c7 > 0 independent of µ. Consider the expression B(x1 ) = ψ −

= −

|i|=2

K+1

µl−2

l=0

i (K) hN = i D ω

|i|=l

i (K) hN −ψ+ i D ω

K

K+1

µl−2

l=3

|i|=l

i (K) . hN i D ω

j Substitute ω (K) (x1 , x2 ) = j=0 µ ωj (x1 , x2 ) and remind (3.1.13),(3.1.18). Then there exist a constant c8 independent of µ such that

|B(x1 )| ≤ c8 µK . Thus |I(φ)| ≤ |

˜µ C

B(x1 )φ dx| + |∆3 (φ)| ≤

≤ c8 µK φH 1 (C˜µ )

mes C˜µ + |∆3 (φ)| ≤

≤ (c8 + c7 )µK φH 1 (C˜µ ) mes C˜µ ≤ ≤ c9 µK φH 1 (C˜µ ) mes C˜µ ,

with a constant c9 > 0 independent of µ. On the other hand, u(K) ∈ H 1 (C˜µ ) and it vanishes when x1 = 0 or x1 = b. Indeed, (K)

u(K) |x1 =0 = (uB =

K+1 l=0

µl

|i|=l

Ni

x

µ

+ Ni0

0(K)

+ uP

x µ

)|x1 =0 + q0 =

|x1 =0 Di ω (K) |x1 =0 + q0 =

145


=

K+1 l=0

µl

|i|=l

0 i (K) hN |x1 =0 + q0 = 0 i D ω

due to relations (3.1.16). Similarly, u(K) |x1 =b = 0. Taking φ = u − u(K) we obtain the inequality κu − u(K) 2H 1 (C˜µ ) ≤ I(u − u(K) ) = =

s

˜µ C m,j=1

Amj

x ∂(u − u(K) ) ∂(u − u(K) )

µ

∂xm

∂xj

≤ c10 µK u − u(K) H 1 (C˜µ ) with a constant c10 > 0 independent of µ. So,

mes C˜µ ,

u − u(K) H 1 (C˜µ ) ≤ c10 /κµK

this completes the proof of the theorem. Corollary 3.1.1. The estimate holds (K−1)

u − uB

0(K−1)

− uP

1(K−1)

− uP

dx ≤ (3.1.23)

mes C˜µ ;

H 1 (C˜µ ) = O(µK ) mes C˜µ , K = 1, 2, ....

1(K) − Indeed, the H 1 −norm of the differences u0(K) − u0(K) χ(x1 ) and u 1 K u χ(b − x1 ) and the H −norm of ρ are of order O(µ ) mes Cµ . Moreover, (K) (K−1) 0(K) 0(K−1) 1(K) 1(K−1) uB − uB H 1 (C˜µ ) , uP − uP H 1 (C˜µ ) , uP − uP H 1˜(Cµ ) = K O(µ ) mes C˜µ . 1(K)

(K−1)

So when one replace in the estimate of the theorem u(K) by uB + 0(K−1) 1(K−1) uP + uP , the order of this estimate does not change. It proves the corollary. (K−1) 0(K−1) 1(K−1) + uP + uP does not vanish at Remark 3.1.3 The function uB K the ends of the rod; however, it has the order O(µ ) there. Remark 3.1.4 The plate can be simulated by a thin cylinder Ωµ = G × (−µ/2, µ/2) ⊂ R3 , where G is a two-dimensional domain with a smooth boundary. In this case the the construction of a complete asymptotic expansion is (0) still an open problem. Nevertheless the estimate for uB can be obtained by means of the technique [16], Chapter 4, section 4.1. Consider equation (3.1.1) set in Ωµ with condition (3.1.2) on the both bases of cylinder Ωµ and (3.1.3) on the lateral boundary of the cylinder ∂G × (−µ/2, µ/2). Let ω0 be a solution of equation (3.1.13) for j = 0, supplied with the Dirichlet condition ω0 = 0 on ∂G. Then the estimate can be proved in the same way as (32)-(33) of Chapter 4, section 4.1 [16]: (0) u − uB H 1 (Ωµ ) = O( µmesΩµ )

146


3.2

Elasticity of a plate.

3.2.1

Statement of the problem.

For s = 2, problem (3.1.1)–(3.1.3) can be regarded as the two-dimensional analog of the elasticity theory system of equations in a plate. Below a three-dimensional formulation (s = 3) will be considered. Consider the elasticity theory system of equations Pu ≡

3 ∂ x ∂u = ψ(x1 , x2 ), Aij µ ∂xj ∂xi i,j=1

µ , x∈C

with boundary conditions 3 x x 1 ∂u ∂u 2 1 = 0, , ,± ≡± A3j 2 ∂xj µ µ ∂ν j=1

x1 = 0

u = 0,

or x1 = b,

µ , x ∈ ∂U

(3.2.1)

(3.2.2) (3.2.3)

and with the condition of T -periodicity with respect to x2 . Here b and T are numbers of the order of 1 that are multiples of µ and Aij (ξ) are 3 × 3 matrices whose elements satisfy the conditions of subsection 2.2.1 and are 1-periodic with respect to ξ1 and ξ2 . These elements are assumed to be piecewise smooth functions, while ψ is a C ∞ three-dimensional vector valued function that is T -periodic with respect to x2 .

3.2.2

Inner expansion

We seek a formal asymptotic solution in the form of a series analogous to the series of the previous section u(∞) =

∞

µl

l=0

Ni

|i|=l

x µ

Di ω(x1 , x2 ),

(3.2.4)

where i = (i1 , . . . , il ) is a multi-index, ij ∈ {1, 2}, ω(x1 , x2 ) is a smooth T periodic with respect to x2 three-dimensional vector function, Ni (ξ) are 3×3 matrix functions that are 1-periodic with respect to ξ1 and ξ2 . Substituting series (3.2.4) in (3.2.1), (3.2.2) and the interface conditions and grouping terms of the same order, we obtain P u(∞) − ψ =

∞ l=0

µl−2

|i|=l

Hi (ξ)Di ω − ψ,

∞ ∂u(∞) = µl−1 Gi (ξ)Di ω, ∂ν l=0

|i|=l

(3.2.5)

(3.2.6)

3.2.

147

ELASTICITY OF A PLATE.

where Hi (ξ) = Lξξ Ni + Ti (ξ), Ti (ξ) =

3 3 ∂N Ni2 ...il ∂ + Ai1 i2 Ni3 ...il , (Aji1 Ni2 ...il ) + Ai1 j ∂ξξj ∂ξ ξ j j=1 j=1

Gi (ξ) =

3 3

m=1

Amj

j=1

∂N Ni + Ami1 Ni2 ...il nm . ∂ξξj

Suppose that Hi (ξ) = hi ,

ξ ∈ IR2 ×

1 1 ; − , 2 2

Gi (ξ) = 0,

1 ξ3 = ± , 2

where hi are constant 3×3 matrices. Assume that [N Ni ] = 0 and [Gi ] = 0 on the surfaces of discontinuity. We obtain the following recurrent chain of problems of the form Lξξ Ni = −T Ti (ξ) + hi , 3

A3j (ξ)

j=1

[N Ni ] Σ = 0,

∂N Ni = −A3i1 Ni2 ...il , ∂ξξj

(3.2.7)

1 ξ3 = ± , 2

- .

3 ∂N

Ni

Ami1 Ni2 ...il

,

= ∂ννξ Σ Σ m=1

(3.2.8)

(3.2.9)

to determine Ni that are 1-periodic functions with respect to ξ1 and ξ2 . The constant matrices hi are chosen from the solvability conditions for problem (3.2.7)-(3.2.9): hi =

3 j=1

Ai1 j

∂N Ni2 ...il + Ai1 i2 Ni3 ...il , ∂ξξj

h∅ = 0,

hi1 = 0;

l ≥ 2, (3.2.10)

here · = (−1/2,1/2)3 dξ and N∅ = I (the identity matrix). The sufficiency of condition (3.2.10) can be proved as in Lemma 2.2.1. Thus, the algorithm for constructing the matrices Ni is recurrent. We formally assume that Ni = 0 for |i| < 0, N∅ = I (the identity matrix), and Ni (ξ) is a solution to problem (3.2.7)–(3.2.10) for l > 0. The right-hand side contains Nj with multi-indices j whose length is smaller than |i|. Let the following condition A′ be satisfied: δ3i +δ3k +δ3j +δ3l kl akl aij (ξ). ij (S3 ξ) = (−1)

(Remind that S3 ξ = (ξ1 , ξ2 , −ξ3 )).

148

CHAPTER 3. HETEROGENEOUS PLATE Theorem 3.2.1. The matrices hi have the form ⎛ 11 ⎞ h12 0 hi i h22 0 ⎠. hi = ⎝h21 i i 0 0 h33 i

kl For |i| = 2, 3 we have h33 i = 0; for |i| = 2 the elements hi1 i2 , k, l ∈ {1, 2}, of the matrices hi satisfy the relations

∀e = (ei1 i2 )i1 ,i2 ∈{1,2} , eli2 = eil 2 , the following inequality holds 2

i1 ,i2 ,k,l=1

l k hkl i1 i2 ei2 ei1 ≥ κ

2

(eli2 )2 ,

i1 l lk hkl i1 i2 = hki2 = hi2 i1 ,

−

2

h33 i1 i2 i3 i4 ηi1 ηi2 ηi3 ηi4 > 0

i1 ,i2 ,i3 ,i4 =1

(3.2.11)

i2 ,l=1

(3.2.12)

∀(η1 , η2 ) = (0, 0).

(3.2.13)

The proof of theorem 3.2.1 will be given later. Then (3.2.5) takes the form P u(∞) − ψ =

2

h i1 i 2

i1 ,i2 =1

+µ2

2 ∂3ω ∂2ω +µ h i1 i 2 i3 ∂xi1 ∂xi2 ∂xi3 ∂xi1 ∂xi2 i ,i ,i =1 1

2

h i 1 i2 i3 i4

i1 ,i2 ,i3 ,i4 =1

+

∞

µl−2

l=5

hi

|i|=l

2

3

∂4ω + ∂xi1 ∂xi2 ∂xi3 ∂xi4

∂lω − ψ = 0. ∂xi1 · · · ∂xil

(3.2.14)

Problem (3.2.14) can be regarded as the homogenized equation of infinite order with respect to the three-dimensional vector ω. A formal asymptotic solution (f.a.s.) of this problem is sought in the form of a series ω=

∞

µj ωj (x1 , x2 ),

(3.2.15)

j=−2

where ωj do not depend on µ and ωj1 = 0 and ωj2 = 0 for j = −2, −1. Substituting the series (3.2.15) into (3.2.14), we obtain a recurrent chain of equations for the components ωjk of the vectors ωj in the form 2

i1 ,i2 =1

A¯i1 i2

∂2ω ˆj = fˆj (x1 , x2 ), ∂xi1 ∂xi2

(3.2.16)

3.2.

149

ELASTICITY OF A PLATE. 2

h33 i1 i 2 i3 i4

i1 ,i2 ,i3 ,i4 =1

∂ 4 ωj3 = fj3 (x1 , x2 ), ∂xi1 ∂xi2 ∂xi3 ∂xi4

(3.2.17)

where ω ˆ j and fˆj are the first two components of the vectors ωj and fj , A¯i1 i2 are ˆ 2 × 2 matrices with components hkl ˆ j1 , j1 < j, i1 i2 , the functions fj depend on ω and on the derivatives of these functions, while the functions fj3 depend on ωj31 , j1 < j. Thus, the f.a.s. of problem (3.2.1), (3.2.2) is constructed.

3.2.3


A formal asymptotic solution of problem (3.2.1)–(3.2.3) will be sought again in the form (2.2.22): u(∞) = uB + u0P + u1P , where uB is defined by (3.2.4) and u0P =

∞ l=0

u1P =

∞ l=0

µl

|i|=l

µl

|i|=l

Ni0 (ξ)Di ω ξ=x/µ ,

Ni1 (ξ)Di ω ξ

1 =(x1 −b)/µ,

ξ ′ =x′ /µ

(3.2.18)

.

(3.2.19)

Here the matrix functions Ni0 (ξ) and Ni1 (ξ) are 1-periodic with respect to ξ2 and exponentially stabilize to zero as ξ1 → +∞, ξ ′ = (ξ2 , ξ3 ), x′ = (x2 , x3 ). Substituting (3.2.18) and (3.2.19) into (3.2.1)–(3.2.3) and taking into account the fact that uB is constructed above, we obtain recurrent chains of problems to determine Ni0 (ξ) as in subsection 3.1.2 1 1 , Lξξ Ni0 + Ti0 (ξ) = 0, ξ1 > 0, ξ ′ ∈ IR × − , 2 2 3 j=1

A3j

∂N Ni0 = −A3i1 Ni02 ...il , ∂ξξj

1 ξ3 = ± , 2

0 Ni (0, ξ ′ ) + Φh Ni0 (0, ξ ′ ) = −N i

(3.2.20)

Ni1 (ξ)

and obtain similar problems for defined for ξ1 < 0 with the constant h1i is a 3×4 matrix instead of h0 . Here Φ i

⎛

1 Φ(ξ) = ⎝0 0

0 1 0

⎞ 0 −ξ3 0 0 ⎠, 1 ξ1

hri are 4×3 constant matrices chosen from the condition that solutions of problems (3.2.20) stabilize to zero as ξ1 → +∞ (ξ1 → −∞ for r = 1), and N∅0 = 0. Theorem 3.2.2. Let Aij (ξ) satisfy the conditions formulated at the beginning of this section, let the three-dimensional vectors Fj (ξ) be 1-periodic with respect to ξ2 and belong to L2 ([0, +∞)×(−1/2, 1/2)2 ), j = 0, 1, 2, 3, let conditions

150


(2.32) be satisfied for β = (−1/2, 1/2)2 , and let a three-dimensional vector u0 (ξ ′ ) be 1-periodic with respect to ξ2 and belong to H 1/2 ({0} × (−1, 1) × (−1/2, 1/2)). Then a solution u(ξ) of the problem Lξξ u = F0 (ξ) +

3 ∂F Fj j=1

3 j=1

A3j

∂u = Fj , ∂ξξj

∂ξξj

,

′

ξ ∈ IR ×

ξ1 > 0,

1 ξ3 = ± , 2

1 1 , − , 2 2

u ξ1 =0 = u0 (ξ ′ )

(3.2.21)

such that u is 1-periodic with respect to ξ2 and inequality (2.3.36) holds for some where h is a constant four-dimensional vector, rigid displacement w(ξ) = Φ(ξ)h, exists and is unique. Here by a solution of problem (3.2.21) we mean a vector function u(ξ) satisfying the integral identity from subsection 3.1.2. Theorem 3.2.2 can be proved similarly to Theorems 4 and 5 in [130]. The r matrices hri are defined similarly to the matrices hN from Chapter 2. In this l case ⎞ ⎛ 1 0 0 ⎜0 1 0⎟ r ⎟ h∅ = ⎜ ⎝0 0 1⎠ 0 0 0

since N∅r = 0 and N∅ = I. The fourth row of the matrix hri1 is equal to (0, 0, 1). The fact that the first two elements of the row are zeros can be proved, as in Lemma 2.3.6, by using the following relation for the components nkl i1 (ξ) of the matrices Ni1 (ξ): δ3k +δ3l kl nkl ni1 (ξ). (3.2.22) i1 (S3 ξ) = (−1)

It follows from this relation that for k, l ∈ {1, 2} these components are even with respect to ξ3 . The number 1 at the end of the row is explained by the fact that the third column of the matrices Ni1 (ξ) has the form (−ξ3 , 0, 0)∗ , and therefore the third column Nir1 is zero and the coefficient for the rigid displacement (−ξ3 , 0, ξ1 ) in boundary condition (3.2.20) is equal to 1. After determining all Nir ’s and hri ’s by induction on |i|, we obtain a homogenized problem for the three-dimensional vector function ω(x1 , x2 ): this problem is formed by equation (3.2.14) with the boundary conditions

ω+

∞

l

µ

l=1

r(cut 3) i hi Dω

|i|=l

= 0, x1 =rb

∞ ∂ω 3

r(oth 3)

µ + µl hi Di ω = 0. ∂x1 x1 =rb l=2

(3.2.23)

|i|=l

Here the superscript ‘cut 3’ denotes the first three rows of the matrix and ‘oth 3’ denotes the fourth row.

3.2.

151


A f.a.s. of problem (3.2.1)–(3.2.3) is sought in the form of series (3.2.15) with boundary conditions

∂ωj3

4 = gjr , r = 0, 1, (3.2.24) ωj x =rb = gjr , r = 0, 1,

1 ∂x1 x1 =rb q where the gjr ’s depend on ωjr1 with j1 < j and on derivatives of these functions. The homogenized problem of zero order is formed by relations (3.2.16), (3.2.17), (3.2.24) with j = 0,

1 , x2 ), f0 (x1 , x2 ) = ψ(x

f03 (x1 , x2 ) = ψ 3 (x1 , x2 ),

g0r = 0.

The estimate for the difference of the exact solution and the asymptotic solution can be obtained as in section 3.1.

3.2.4

Proof of Theorem 3.2.1.

As in Lemma 2.3.2, by substituting into (3.2.7)–(3.2.10), we can prove the relation δ3k +δ3l kl nkl ni (ξ); i (S3 ξ) = (−1) kl for the elements hkl i of the matrices hi this relation implies the formula hi = δ3k +δ3l kl kl (−1) hi which yields hi = 0 for k = 3 and l = 3 or for k = 3 and l = 3. It follows from (3.2.10) that

hi 1 i 2 =

3

Ai1 j

3 ∂ ∂N N i2 (N Ni2 + Iξi2 ) + Ai1 i2 = Ai 1 j ∂ξξj ∂ξξj j=1

Amj

∂ ∂ (N Ni1 + Iξi1 ) . (N Ni2 + Iξi2 ) ∂ξm ∂ξξj

j=1

=

3

m,j=1

(3.2.25)

The last relation arises from the integral identity for problem (3.2.7)–(3.2.9) with |i| = 2. Relations (3.2.11) and (3.2.12) can be derived from (3.2.25). Indeed, hkl i1 i 2 =

3

m,j,q,p=1

=

3

m,j,q,p=1

where

Zjpl (N Nr + Iξr ) =

aqp mj

∂ ∂ (nqk (npl i1 + ξi1 δqk ) i2 + ξi2 δpl ) ∂ξm ∂ξξj

pl qk aqp Z (N N + Iξ )Z Z (N N + Iξ ) , i2 i2 i1 i1 m mj j

∂ 1 ∂ (njl (npl r + ξr δjl ) . r + ξr δpl ) + ∂ξξp 2 ∂ξξj

This implies (3.2.12). For ηik1 = ηki1 we have 2

i1 ,i2 ,k,l=1

l k hkl i1 i2 ηi2 ηi1 =

3

2

m,j,q,p=1 i1 ,i2 ,k,l=1

pl qk aqp Ni2 +Iξi2 )ηil2 Zm (N Ni1 +Iξi1 )ηik1 mj Zj (N

152


≥κ

3 2

j,p=1

≥κ =κ

2

j,p=1

1

2

Zjpl (N Ni2

i2 ,l=1

2 2

j,p=1

2

+

Iξi2 )ηip2

Zjpl (Iξi2 )ηip2

i2 ,l=1

(δδpl δji2 + δjl δpi2 )ηip2

i2 ,l=1

2

2

2

=κ

2

(ηjp )2 .

j,p=1

The third column of the matrices Ni1 is (−ξ3 , 0, 0)∗ , hence, nq3 i1 + ξi1 δq3 is a rigid displacement and h33 i1 i2 = 0; moreover,

=

∂ p3 i1 p p3 3p p3 3p ∂ ni2 i3 + a3i n np3 = ai3j1 p h33 i1 i2 i3 = ai1 j i2 i3 + ai1 i2 ni3 2 i3 ∂ξξj ∂ξξj 1p aimj

=

∂ ∂ξ ∂ p3 ∂ p3 3 i1 p p3 1p 1p ni2 i3 + ami n ξ3 =− aimj ni2 i3 + aimi np3 2 i3 2 i3 ∂ξξj ∂ξm ∂ξm ∂ξξj

aii12 pj

∂np3 ∂ i3 (np3 + aii12 pi3 δp3 ξ3 = aii12 pj i3 + δp3 ξi3 ) ξ3 = 0. ∂ξξj ∂ξξj

The proof of the fact that the form (3.2.13) is positive definite is similar to the proof of Lemmas 2.3.4 and 2.3.5. Namely, the following statement can be proved as in Lemma 2.3.4. Let η1 , η2 be two positive numbers. Let operators (2.3.93) and (2.3.94) be defined µ and x ∈ ∂ U µ , respectively. Denote for x ∈ U π π µ µ π π . × − , × − , Cµη = − , 2 2 η2 η2 η1 η1

Assume that a three-dimensional vector function u ∈ H 1π 1

(C Cµη )−norm)

η1

, ηπ −per ,

where H 1π

2

η1

, ηπ −per 2 2

is a completion (by the H of the space of differentiable in IR × [− µ2 , µ2 ] three-dimensional vector functions, 2π/η1 -periodic with respect to x1 and 2π/η2 -periodic with respect to x2 . Assume that u dx = 0. η Cµ

Then the following inequality holds (for sufficiently small µ): 3

i,j=1

η Cµ

Aij

∂u ∂u dx ≥ c µ4 ∇u2L2 (Cµη ) + µ2 u2L2 (Cµη ) , , ∂xj ∂xi

c > 0,

(3.2.26) c1 does not depend on µ. Indeed, in (2.3.93) and (2.3.94) let us perform the following changes of variables:

3.2.

153


1) x1 = ξ1 , x2 = ξ2 , x3 = ξ3 µ, 2) u1 = w1 , u2 = w2 , u3 = w3 /µ, 3) multiply the third component of (2.3.93) and the first two components of (2.3.94) by µ−1 and the last component of (2.3.94) by µ−2 ; then we obtain for (2.3.93) and (2.3.94)the following presentations: −

3 ∂w ∂ , Bij (ξ) ∂ξξj ∂ξi i,j=1

ξ3 ∈

1 1 , − , 2 2

3

B1j (ξ)

j=1

∂w , ∂ξξj

1 ξ3 = ± , 2

where the elements bkl ij of the matrices Bij are δi3 +δj3 +δk3 +δl3 kl bkl . ij (ξ) = aij ξ1 /µ, ξ2 /µ, ξ3 1/µ

As in Lemma 2.3.6, we see that for any symmetric matrix with elements ηik the following inequality holds: 3

i,j,k,l=1

and we have w dξ = 0, Qη 0

k l bkl ij ηi ηj ≥ κ

3

(ηjl )2

j,l=1

Qη0 = − π/η1 , π/η1 × − π/η2 , π/η2 × − 12 , 12 .

where

Korn’s inequality for the function w implies 3

i,j=1

Qη 0

Bij

∂w ∂w dξ ≥ κc1 w2H 1 (Qη ) , , 0 ∂ξξj ∂ξi

where c1 is a positive constant independent of µ. The left-hand side of the last inequality is equal to the left-hand side of inequality (3.2.26) multiplied by µ−1 and we have w2H 1 (Qη ) ≥ µ3 ∇u2L2 (Cµη ) + µu2L2 (Cµη ) , 0

which implies (3.2.26). µ we specify a vector function Furthermore, in U u(K) µ (x) =

K+1

µl

l=0

|i|=l

Ni

x

µ

Di V (x1 , x2 ),

where Ni (ξ) are the above matrices and V (x1 , x2 ) are infinitely differentiable three-dimensional vector functions that are 2π/ηj -periodic with respect to xj , j = 1, 2, and the first two components of V are zero: V = (0, 0, V 3 )∗ . As in Lemma 2.3.5, for sufficiently large K and given ηj , we obtain 3

i,j=1

η Cµ

Aij

(K) (K) ∂uµ ∂uµ dx , ∂xi ∂xj

154


= −µ3

2

h33 i1 i2 i3 i 4

i1 ,i2 ,i3 ,i4 =1

2 η ≥ ≥ µ2 κc1 u(K) µ L2 (Cµ )

(Di V 3 )V 3 dx1 dx2 + O(µ4 )

(−π/η1 ,π/η1 )×(−π/η2 ,π/η2 )

µ3 κc1 2

(V 3 )2 dx1 dx2 ,

(−π/η1 ,π/η1 )×(−π/η2 ,π/η2 )

c1 being constant independent of µ. By setting V 3 = sin(η1 x1 + η2 x2 ) we see that Di V = ηi1 ηi2 ηi3 ηi4 V 3 ; this implies −

2

i1 ,i2 ,i3 ,i4 =1

hi33 η η η η ≥ 1 i 2 i 3 i 4 i1 i2 i 3 i 4

κc1 >0 2

for all µ small enough. Remind that here η1 > 0 and η2 > 0 are arbitrary positive numbers. Considering V 3 = sin(±η1 x1 ± η2 x2 ) we can generalize this inequality for any couples (η1 , η2 ) = (0, 0). (If η1 = 0 and η2 = 0 or η2 = 0 and η1 = 0 we apply the same reasoning with Cµη = (− 12 , 12 ) × (− ηπ2 , ηπ2 ) × (− µ2 , µ2 ) or Cµη = (− ηπ1 , ηπ1 ) × (− 12 , 12 ) × (− µ2 , µ2 ) respectively). The proof of Theorem 3.2.1 is complete.

3.2.5

Algorithm for calculating the effective stiffness of a plate

As in section 2.3, we present a calculation algorithm for the coefficients hkl i1 i 2 , i1 , i2 , k, l ∈ {1, 2}, of the homogenized equation (elasticity modules in a plane) kl and the coefficients h33 ¯kl i1 i2 i3 i4 , i1 , i2 , i3 , i4 ∈ {1, 2}. Below we denote a i1 i2 = hi1 i2 33 and ¯i1 i2 i3 i4 = −hi1 i2 i3 i4 (bending stiffness coefficients). We first solve the elasticity theory systems of equations for i1 , q = 1, 2:

∂ ∂ (q) (N Ni1 + eq ξi1 ) = 0, Amj ∂ξξj ∂ξm m,j=1 3

3 j=1

A3j

∂ (q) (N N + eq ξi1 ) = 0, ∂ξξj i1

ξ3 ∈

1 1 , − , 2 2

1 ξ3 = ± , 2

(q)

where eq = (δq1 , δq2 , δq3 )∗ and Ni1 is an unknown three-dimensional vector function that is 1-periodic with respect to ξ1 and ξ2 . In this case the elasticity modules in a plane can be defined by the following formulas:

kr ∂ (nrq a ¯kq i2 + δrq ξi1 ) . i1 i2 = ai1 j ∂ξξj Then we solve the elasticity theory systems of equations for i1 , i2 = 1, 2:

∂ ∂ (3) Ni1 i2 − Ami1 ei2 ξ3 = 0, Amj ∂ξξj ∂ξm m,j=1 3

ξ3 ∈

1 1 , − , 2 2

3.2.

155

ELASTICITY OF A PLATE. 3

A3j

j=1

∂ (3) N − A3i1 ei2 ξ3 = 0, ∂ξξj i1 i2

1 ξ3 = ± , 2

(3)

related to three-dimensional vector functions Ni1 i2 (ξ) that are 1-periodic in ξ1 and ξ2 . The bending stiffness coefficients ¯i1 i2 i3 i4 are defined by the formula

3 ∂nr3 . c¯i1 i2 i3 i4 = − ξ3 aii12 rj i3 i4 − aii12 ii43 ξ3 ∂ξξj j,r=1 Indeed

i1 p p3 3p ∂ 3p p3 i1 p ∂ np3 h33 np3 i1 i2 i3 i4 = ai1 j i2 i3 i4 + a3i2 ni3 i4 i2 i3 i4 + ai1 i2 ni3 i4 = a3j ∂ξξj ∂ξξj

=

1p aimj

=

∂ ∂ξ ∂ p3 ∂ p3 3 1p 1p 1p ni2 i3 i4 +aimi np3 ξ3 aimj =− ni2 i3 i4 +aimi np3 2 i3 i4 2 i3 i 4 ∂ξξj ∂ξm ∂ξm ∂ξξj

aii12 pj

p3 ∂np3 i3 i4 i1 p ∂ni3 i4 + aii12 pi3 (−ξ3 δpi4 ) ξ3 + aii12 pi3 np3 = ξ a 3 i4 i2 j ∂ξξj ∂ξξj ∂ p3 ni3 i4 − aii12 ii43 ξ3 ξ3 . = aii12 pj ∂ξξj

For constant coefficients, in the isotropic case we have: a ¯11 ¯22 11 = a 22 =

c¯1111 = c¯2222 =

¯ (3) E , 1 − ν2

¯ (3) E , 12(1 − ν 2 )

a ¯11 22 = M ,

12 ¯a12 =

c¯1212 = M /12 ,

¯ (3 ) νE , 1 − ν2

¯c1122 =

¯ (3 ) νE , 12 (1 − ν 2 )

where ν = λ/(2(λ + M )). Consider an orthotropic case in which for each the tensor akl ij (ξ) of elastic modules is given by means of nine scalar functions: E1 (ξ), E2 (ξ), E3 (ξ) (Young modules), µ21 (ξ), µ31 (ξ), µ32 (ξ) (shear modules), and ν21 (ξ), ν31 (ξ), ν32 (ξ) (Poisson ratios) as follows: ⎛ 11 a11 ⎝a12 12 a13 13

a12 12 a22 22 a23 23

⎛ 1 ⎞ ⎜ E1 a13 13 ⎜ ν21 23 ⎠ a23 = ⎜ ⎜ − E2 ⎝ ν a33 33 31 − E3

ν21 E2 1 E2 ν32 − E3

−

ν31 ⎞−1 E3 ⎟ ν32 ⎟ ⎟ , − E3 ⎟ 1 ⎠

−

E3

33 33 a22 11 = µ21 , a11 = µ31 , a22 = µ32 , and if among the indices i, j, k, l there is an kl index that differs from the others, then akl ij = 0. Let aij depend on ξ3 only. Then we have 1 1 22 11 a ¯11 , a ¯22 = a ¯11 = 22 = µ21 , 2 /E 2 ) , 2 /E E1 (1/(E1 E2 ) − ν21 1/E1 − ν21 2 2

156

CHAPTER 3. HETEROGENEOUS PLATE a ¯12 12

ξ32 ν21 , , c¯1111 = = 2 /E 2 /E 1/E1 − ν21 1/E1 − ν21 2 2 ξ32 c¯2222 = 2 /E 2 ) , E1 (1/(E1 E2 ) − ν21 2 ν21 2 c¯ 1122 = 2 2 /E + 2µ21 ξ3 , 1/E1 − ν21 2

(3.2.27)

1/2 where · = −1/2 dξ3 ; the last coefficient is the sum of coefficients with all possible permutations of subscripts. If among the indices of an element a ¯kl ij there is an index that differs from kl the others, then a ¯ij = 0. The sums of coefficients ci1 i2 i3 i4 over all possible permutations of the sets (1, 2, 2, 2) and (2, 1, 1, 1) are equal to zero.

3.3

Equivalent homogeneous plate problem

Here below we discuss the problem of existence of a homogeneous orthotropic plate that is mechanically equivalent to the given heterogeneous locally orthotropic plate. Besides the nine elasticity modules we can variate the thickness of the plate. Here we will find out that in general case an equivalent homogeneous plate does not exist (remind that it always exists for conductivity equation, compare to section 3.1). We formulate the necessary and sufficient conditions of existence of an equivalent homogeneous orthotropic plate and propose the approximating homogeneous plate when these conditions are not true. 3 akl In mechanics it is usual to take µ¯ ¯i1 i2 i3 i4 as the bending characij and µ c teristic and the membrane characteristic of the stiffness of plates, where µ is the thickness of a plate.

Figure 3.3.1. The heterogeneous plate and an equivalent homogeneous plate

157

3.3. EQUIVALENT HOMOGENEOUS PLATE PROBLEM The homogenized problem of zero order for a plate has the form 2

i,j,k,l=1

c¯1111

∂ kl ∂ω0l = ψ k (x1 , x2 ), a ¯ ∂xi ij ∂xj

∂ 4 ω3 ∂ 4 ω03 ∂ 4 ω03 = ψ 3 (x1 , x2 ), + ¯1122 2 0 2 + ¯2222 4 ∂x42 ∂x1 ∂x1 ∂x2

x1 ∈ (0, b), ω0 = 0,

∂ω03

x2 ∈ IR,

k = 1, 2,

= 0 for x1 = 0, b. ∂x1 Now we consider the following existence problem for a plate that is homogeneous (with respect to plane and bending characteristics) and equivalent to a kl stratified plate with the characteristics µ¯ aij and µ3 c¯ijkl defined by (3.2.27); i.e., ˆ1 , for the eight given characteristics (3.2.27) we seek nine constant modules E ˆ ˆ E 2 , E3 , µ ˆ21 , µ ˆ31 , µ ˆ32 , νˆ21 , νˆ31 , and νˆ32 and a new thickness of the equivalent plate µH (thus, we have ten unknowns in all) such that the plane characteris3 tics (µH)ˆ akl îjkl corresponding to these ten ij and bending characteristics (µH) c 3 ¯ijkl respectively: values coincide with the given characteristics µ¯ akl ij and µ c (µH)ˆ akl akl ij = µ¯ ij ,

(µH)3 cîjkl = µ3 c¯ijkl

(3.3.1)

Clearly, we can calculate the values a ˆkl îjkl from (3.2.27) by replacing ij and c ˆp , µ the modules Ep , µpq , and νpq by unknown constants( E ) ˆpq , and νˆpq and by replacing the brackets by 1/12 (for example, we set ξ32 = 1/12 ). Note that (3.2.27) does not depend on E3 , µ31 , µ32 , ν31 , and ν32 . Thus, from (3.3.1) we ˆ1 , E ˆ2 , µ ˆ21 , νˆ21 , pass to the following equivalent problem: to find five constants E and H that satisfy the eight equations

1 a ¯11 11 , = 2 /E ˆ ˆ2 H 1/E1 − νˆ21

a ¯11 22 = µ21 , H

1 c¯1111 , = 3 2 /E ˆ ˆ2 ) H 12(1/E1 − νˆ21

µ ˆ21 c¯1212 , = 12 H3

a ¯22 1 22 , = 2 /E ˆ ˆ ˆ ˆ2) H E1 (1/(E1 E2 ) − νˆ21 2

νˆ21 a ¯12 12 , = 2 /E ˆ ˆ2 H 1/E1 − νˆ21

(3.3.2)

1 c¯2222 , = 3 2 /E ˆ ˆ ˆ ˆ2) H 12E1 (1/(E1 E2 ) − νˆ21 2

νˆ21 c¯1122 , = ˆ1 − νˆ2 /E ˆ2 ) H3 12(1/E 21

(3.3.3)

a ¯12 c¯1212 . = 12 a ¯11 c¯1111 11

(3.3.4)

with given a ¯11 ¯22 ¯11 ¯12 ¯2222 , c¯1122 , c¯1212 . 11 , a 22 , a 22 , a 12 and ¯1111 , c Now we can state necessary and sufficient conditions for system (3.3.1) to be solvable (and hence existence conditions for equivalent homogeneous plate):

a ¯22 c¯2222 , = 22 a ¯11 c¯1111 11

a ¯11 c¯1122 , = 22 a ¯11 c¯1111 11

158


If conditions (3.3.4) hold, then five constants for equivalent plate can be found from the following relations: c1111 /¯ a11 µ ˆ21 = a ¯11 νˆ21 = a ¯12 a11 H = 12¯ 22 /H, 12 /¯ 11 , 11 , 2 ˆ2 = (¯ E a22 ¯11 ˆ21 )/H, 22 − a 11 ν

2 ˆ1 = a ˆ ˆ E ¯11 ¯11 ˆ21 ). 11 E2 /(H E2 + a 11 ν

(3.3.5)

Indeed, relation (3.3.4) holds because the right-hand sides of (3.3.2) and (3.3.3) are proportional. We find H by dividing the first equation of (3.3.2) by the first equation of (3.3.3); then we can find µ ˆ21 from the third equation of 3.3.2) and determine νˆ21 as the ratio by dividing the fourth equation of (3.3.2) ˆ1 and E ˆ2 from the first two by the first equation of (3.3.2). Finally, we obtain E equations of (3.3.2). Relations (3.3.3) follow from (3.3.2) and (3.3.4). This result was obtained in cooperation with V. E. Grebennikov. Conditions (3.3.4) hold in particular if the mono-layers are isotropic and symmetric. If conditions (3.3.4) fail, then the equivalent plate does not exist. Nevertheless, one can apply relations (3.3.5) to find an approximation of the five unknowns. The error of this approximation can be evaluated by ∆=

c¯

2222

c¯1111

−

2 c¯ a ¯12 2 a ¯11 2 c¯1212 a ¯22 1122 22 . + − 12 + − 22 11 11 a ¯11 a ¯11 c¯1111 a ¯11 c¯1111 11

This approach was used to adopt a finite element method software based on the concept of a homogeneous plane finite element to the analysis of thin-walled constructions of stratified composite materials (for example, of a stratified car body).

3.4

Time dependent elasticity problem for a plate.

Consider a non-stationary elasticity equation −ρ(x/µ)

3 ∂u ∂ ∂2u = ψµ (x1 , x2 , t), (x/µ) A + mj ∂xj ∂t2 m,j=1 ∂xm

x ∈ Cµ , (3.4.1)

with boundary conditions (3.2.2), (3.2.3), junction conditions (2.3.4), and initial conditions

∂u

= 0. (3.4.2) u t=0 = 0,

∂t t=0

Here Aij (ξ) are matrix functions from section 3.2 and the function ρ(ξ) is 1periodic with respect to ξ1 , ρ(ξ) ≥ κ > 0, and ρ(ξ) is piecewise smooth in the sense of section 2.3. Besides, ρ(SA ξ) = ρ(ξ) for A = 2, 3 and ψµ (x, t) = (ψ 1 (x1 , x2 , t), ψ 2 (x1 , x2 , t), µ2 ψ 3 (x1 , x2 , µt))∗ , ψ k ∈ C ∞ , k = 1, 2, ψ 3 (x1 , x2 , θ) ∈ C ∞ ,

ψ 3 (x1 , x2 , θ) = 0 for θ ≤ θ0 ,

θ0 > 0.

159

3.4. TIME DEPENDENT ELASTICITY PROBLEM FOR A PLATE.

The inner expansion of solution to problem (3.4.1), (3.4.2), (3.2.2), (3.2.3), (2.3.4) is sought as a series u(∞) =

∞

µl+q Niq (x/µ)

l+q=0 |i|=l, i=(i1 ,...,il ),ij ∈{1,2}

∂q i D ω(x1 , x2 , t) ∂tq

(3.4.3)

where ω is a three-dimensional vector of the form (ω 1 (x1 , x2 , t), ω 2 (x1 , x2 , t), ω 3 (x1 , x2 , µt))∗ , Niq are analogs of the functions Niq from section 3.2; Nlq are (1-periodic with respect to ξ1 ) solutions of the problems N Lξξ Niq = −T Tiq (ξ) + hN iq ,

ξ ∈ IR2 × (−1/2, 1/2),

∂N Niq = −A1i1 Ni2 ,...,il ,q , ∂ννξ

[N Niq ] Σ = 0, N Tiq (ξ) =

-

ξ3 = ±1/2,

.

3

∂N Niq

= − A N n

, mi i ,...,i ,q m 1 2 l ∂ννξ Σ Σ m=1

(3.4.4)

(3.4.5)

(3.4.6)

3 3 ∂ ∂N Ni2 ,...,il ,q +Ai1 i2 Ni3 ,...,il ,q −ρN Ni,q−2 , (Aji1 Ni2 ,...,il ,q )+ Ai1 j ∂ξξj ∂ξ ξ j j=1 j=1

hN iq =

3

Ai1 j

j=1

∂N Ni2 ,...,il ,q ∂ξξj

+ Ai1 i2 Ni3 ,...,il ,q − ρN Ni,q−2 .

(3.4.7) (3.4.8)

As in section 2.3, we obtain a homogenized equation ∞

µl+q−2 hN iq

l+q=2 |i|=l, ij ∈{1,2}

∂q Di ω − ψµ = 0, ∂xl1 ∂tq

(3.4.9)

whose principal part has the form −ρ

∂ 2 ωr + ∂t2

2

q,j,k,r=1

r ∂ kr ∂ω ) − ψ k (x1 , t) + · · · = 0, k = 1, 2, (¯ aqj ∂xj ∂xq

2 ∂ω 3 ∂2 ∂ 2 ω3 3 )−ψ (x , θ)+· · · = 0, (3.4.10) − (¯ µ2 −ρ 1 qjpr ∂xp ∂xr ∂θ2 q,j,p,r=1 ∂xq ∂xj where θ = µt, and the homogenized coefficients a ¯kr qj and ¯qjpr are defined above in section 3.2.

160


Let us seek f.a.s. ω of (3.4.9) in the form of a regular series in powers of µ: ω=

∞

µj ωj .

j=0

The boundary layer corrector can be constructed similarly to that in section 3.2; the same is true for the justification of the asymptotic expansion: it is similar to that of subsection 2.4.4; the a priori estimate follows from [16], [55], [129].

3.5


The asymptotic analysis of the elasticity equations for homogeneous plates (without strict mathematical justification) was first carried out by A.L.Goldenveizer [59] - [61], and later with mathematical justification by Ph.Ciarlet and P.Destuynder [40], B.Lidsky and S.Nazarov [111]. The review of the results on the asymptotic reduction of dimension from 3D to 2D in the plate theory is given in [38], [39] (see, for example, the assessment in the end of Chapter 3 [38]. For an inhomogeneous plate, the homogenized equation of zero order was obtained by D.Caillerie [31]. As in the general homogenization theory, there are two principal tools in the reduction of dimension from 3D to 2D: the convergence techniques and the asymptotic expansions; as it was stated above, the expansions give more information about the structure of the solution but this technique demands more regularity of the data. The complete asymptotic expansion of a solution of the elasticity theory system of equations for an inhomogeneous plate (taking into account the boundary layers) was constructed by G.Panasenko and M.Reztsov [165] and also in [138],[139],[150]. The plates with rapidly varying thickness were considered in [72],[73]and later in [69]. The singular data models are studied in [191]. We do not discuss in the present book the dimensional reduction for non-linear elasticity, for the shell theory. the review on the state of these questions can be found in [38], [110],[176],[179] and in [97].

Chapter 4

Finite Rod Structures We consider finite rod structures, i.e. finite connected unions of rods. The definition of the considered geometrical model of such structures is given in section 4.1. We introduce the notions of L-convergence and of FL-convergence for such structures. This tool gives the main term of solution to conductivity problem in a finite rod structure. This approach is implemented then to elasticity equation and it is applied to a shape optimization problem of minimal compliance (section 4.2). Further we develop a more refined analysis of fields in finite rod structures including asymptotic expansion of elasticity equation based on the expansions of Chapter 2. Namely, these expansions of Chapter 2 are applied to each bar of finite rod structure and then the boundary layers are constructed in neighborhoods of ends. These boundary layers match the inner expansions of each rod and they are similar to that of section 2.3. Finally, the Stokes and the Navier-Stokes equations are studied in a finite system of tubes (or pipes) that is also a variety of finite rod structures.

4.1

Definitions. L-convergence.

Below we consider the finite rod structures; for example , in the two-dimensional α case the finite rod structure is a connected union of thin rectangles Bh,j,µ . Their thickness has a magnitude µ 0 Bh,j,µ . The ends of these segments are called nodes. Then we consider the boundary value problem for a nonlinear partial derivative equation stated on a finite rod structure. The asymptotic reduction of the stated problem is done. The main mathematical result of the section is the passage to the limit in the boundary value problem as µ tends to zero. The formal procedure of the reduction of the initial problem to a system of algebraic equations is justified in the sense of so called FL-convergence (i.e. convergence of the normalized Dirichlet’s integral). It is

161

162

CHAPTER 4. FINITE ROD STRUCTURES

the statement of Theorem 4.1.1. In the case of a scalar elliptic linear problem FL-convergence implies L-convergence, i.e. convergence in normalized L2 norm. The proof of Theorem 4.1.1 uses some auxiliary inequalities (in particular the Poincare´ - Friedrichs inequality for a finite rod structure). These inequalities are proved in Appendix A4.1 and Appendix A4.2.

4.1.1

Finite rod structure

First we define some notions [148]: finite rod structure, nodes, skeleton, sections, nodal domain. Let β1 , ..., βJ be bounded domains in IRs−1 , (s = 2, 3) with a piecewise smooth boundary, Bj,µ (j = 1, ..., J) the cylinders, defined by Bj,µ = {x ∈ Rs | (x2 /µ, ..., xs /µ) ∈ βj , x1 ∈ IR}, α ˜ with and Bh,j,µ the cylinder obtained from Bj,µ by orthogonal transformation Π T T α the matrix α , α = (αil ) and a translation h = (h1 , ..., hs ) . Let eh be an ˜ and h from the vector i(h, α) of s-dimensional vector obtained by means of Π the axis Ox1 with the beginning at the point O . Definition 4.1.1. Let B be a union of all segments eα h when α belongs to a 2 set ∆ ⊂ IRs and h to a set Hα ⊂ IRs , and these sets are independent of µ. Let B be such that any two segments eα h can have only one common point which is the end point for both segments. The set B is called skeleton, the end points of eα h are called nodes. If x0 is an endpoint for a segment e we say that e is an initial segment for x0 . α ˜α We associate the cylinder Bh,j,µ with every eα h and denote by Bh,j,µ the part α of Bh,j,µ enclosed between the two planes passing through the ends of segment ˜α eα h and perpendicular to it (we assume that the bases belong to Bh,j,µ ). Suppose that ∆ and Hα are finite sets and B is connected. Definition 4.1.2. By a finite rod structure we understand the set of interior points of the union

Bµ = Uα∈∆

Uh∈Hα

˜α . B h,j,µ

We assume that Bµ is connected, satisfies the strong cone condition [55] and that µ is a small parameter. Definition 4.1.3. By sections of a rod structure Bµ we understand the ˜α cylinders that are the maximal subsets of the cylinders B h,j,µ for which any ˜α cross-section by the plane perpendicular to the generatrix of B h,j,µ is free of points of other cylinders . We denote by S0 the union of sections. Let c0 µ ≥ d, where d is the maximal diameter of the connected subsets Bµ \S0 , with c0 being independent of µ.

4.1. DEFINITIONS. L-CONVERGENCE.

163

˜ α,+ (respectively B ˜ α,++ ) be a part of B ˜ α , contained between the Let B h,j,µ h,j,µ h,j,µ ˜ α . Let planes spaced by c0 µ (respectively by (c0 + 1)µ) from the bases of B h,j,µ α,+ α,++ ++ ˜ ˜ Bµ+ be a union of all B be a union of all B h,j,µ and let Bµ h,j,µ . ¯µ++ is called the nodal domain. Definition 4.1.4. The domain Bµ \B Remark 4.1.1. We shall also consider as a finite rod structure the union of the set Bµ with some s-dimensional cubes with the edge d0 µ = 2max{d , µ maxj=1,...,J diam βj }

√ and the centers at some nodes. Then we suppose that c0 > 2 sd0 .

Figure 4.1.1. A skeleton B containing three segments

164


Figure 4.1.2. The corresponding finite rod structure Bµ .

Figure 4.1.3. Sections S0


165

˜ α,+ and B ˜ α,++ Figure 4.1.4. Sets B h,j,µ h,j,µ

4.1.2

L-convergence method for a finite rod structure

Here we consider finite rod structures described in Remark 4.1.1 stretched between the planes {x1 = 0} and {x1 = 1} with the loading at the intersection of the finite rod structure with the plane {x1 = 1} by a proper distribution of the material. Let G0 , G1 be two (s-1)-dimensional domains belonging to the planes {x1 = 0} and {x1 = 1} respectively. We consider the set Q of finite rod structures intersecting the planes {x1 = 0} and {x1 = 1} such that 1) the skeleton B belongs to a cube [0, 1]s ; 2) the sets of nodes belonging to the domains G0 ⊂ {x1 = 0} and G1 ⊂ {x1 = 1} are not empty; 3) the rod structure Bµ has additional cubes of Remark 4.1.1 Cx0 = (x01 − d0 µ , x01 + d0 µ) × ... × (x0s − d0 µ , x0s + d0 µ) for all nodes x0 = (x01 , ..., x0s ) from G0 ∪ G1 . We suppose that the constant d0 is such that the rod structure without these additional cubes does not intersect the planes {x1 = −d0 µ} and {x1 = 1 + d0 µ}. These assumptions are not crucial; they are introduced for simplicity of description of the method and its application to the shape design. Denote Σ0 the union of the sides of the cubes Cx0 belonging to the plane {x1 = −d0 µ} and Σ1 the union of the sides of the cubes Cx0 belonging to the plane {x1 = 1 + d0 µ}.

166


Figure 4.1.5. Rod structure with the supplementary cubes Consider the problem, set in a finite rod structure Bµ ∈ Q for the fixed (given) vector-valued constant T : s i=1

∂uµ /∂ν =

s i=1

∂uµ /∂ν =

s i=1

∂/∂xi ( Ai (∇x uµ )) = 0,

f or x ∈ Bµ ,

ni Ai (∇x uµ ) = 0, f or x ∈ Σ = ∂Bµ \(Σ0 ∪ Σ1 ),

(4.1.1)

(4.1.2)

ni Ai (∇x uµ ) = t = µs−1 T /measΣ1 , f or x ∈ Σ1 , (4.1.3) uµ = 0, f or x ∈ Σ0 ,

(4.1.4)

where T, is a constant s−dimensional vector uµ (x), Ai (y) (i = 1, ..., s) are n−dimensional vector-valued functions, ∇x u = (∂uk /∂xl ) is an n × s matrix, y = (ykl ) is an n × s matrix-argument, Ai are assumed to be continuously differentiable, and ni (x) is the cosine of the angle between the axis Oxi and the exterior normal vector n(x). We suppose that the problem (4.1.1)-(4.1.4) has the solution, such that for each n−dimensional vector-valued function ϕ(x) which belongs to the space H 1 (Bµ ) and equal to zero on the surface Σ0 , the integral

167


Bµ

is defined and is equal to

s

(Ai (∇uµ ) , ∂ϕ/∂xi ) dx

i=1

(t , ϕ) dx.

Σ1

Remark 4.1.2. In particular, one can consider Ai (∇x u) =

s

Aij

j=1

∂u , ∂xj

where Aij are constants such that the matrix (Aij )1≤i,j≤s is positive definite and symmetric. In this linear case problem (4.1.1)-(4.1.4) surely has a unique solution. We develop the asymptotic reduction of the problem as µ tends to zero. Here we introduce two types of ”convergence” : the first is L-convergence, i.e. convergence in the normalized L2 (Bµ ) norm. The necessity of normalization is explained by the small measure of the domain Bµ ( meas Bµ tends to zero). The second type of ”convergence” is FL-convergence, i.e. convergence of specially normalized Dirichlet integral. We describe the formal algorithm of construction of the asymptotic solution (1) uµ of the problem (4.1.1) - (4.1.4). The main theorem is proved; it determines the FL-convergence of the solution uµ of the problem (4.1.1) - (4.1.4) ”to” the (1) asymptotic solution uµ of the problem (4.1.1) - (4.1.4). In the linear case the FL-convergence implies L-convergence. Definition 4.1.5. Let uµ (x) be a sequence of functions from L2 (Bµ ) , u0 (x) ∈ L2 (B ); (so u0 is a function defined on the skeleton B ). One says that uµ L - converges to u0 on Bµ if and only if uµ − u ˜0 L2 (Bµ ) → 0, meas(Bµ )

(µ → 0),

where u ˜0 is an extension of u0 (x) onto Bµ such that the values at each point x0 of the set Bµ+ is equal to the value of u0 at the orthogonal projection of x0 on the corresponding segment of B, and at each connected component of a domain Bµ \Bµ+ we pose u ˜0 equal to its value at the node. The normalization factor 1/ meas(Bµ ) is necessary because meas(Bµ ). Notice that L-convergence is not a convergence 1L2 (Bµ ) = in common sense because the domain depends on small parameters. Denote H1,0 (Bµ ) = {ϕ ∈ H1 (Bµ ), ϕ|Σ0 = 0} and

u1 ; u2 F L(Bµ ) =

168


6 * s | Bµ i=1 (Ai (∇u1 ) − Ai (∇u2 ) , ∂ϕ/∂xi ) dx| . supϕ∈H1,0 (Bµ ) ϕH 1 (Bµ ) meas(Bµ )

(4.1.5)

Definition 4.1.6. Let u1,µ (x) and u2,µ (x) be two sequences of functions from H1,0 (Bµ ) . One says that the pair (u1,µ ; u2,µ ) FL - converges if and only if u1,µ ; u2,µ F L(Bµ ) → 0,

(µ → 0).

(4.1.6)

Notice that u1,µ ; u2,µ F L(Bµ ) is not a norm in a general case. Nevertheless in the linear case for the coercive operator (1) it is a norm of the difference of two functions. The asymptotic reduction of the problem (4.1.1)-(4.1.4) analogous to that of [134], [136] and to FL-convergence techniques from [148] deduces it to the algebraic system of equations. To each segment e ⊂ B , (e = eα h )we assign a collection of n−dimensional vectors X0 , ..., Xs , which satisfies the following three conditions: 1) s Ai (X1 , ..., Xs ) νij = 0, (4.1.7) i=1

where j = 1, ..., s − 1, and the vectors

ν 1 = (ν11 , ..., νs1 ) , ... , ν s−1 = (ν1s−1 , ..., νss−1 ) form the basis in (s − 1)−dimensional space orthogonal to e. 2) Let the segments e1 , ..., eq have a common end point; if this end point does not belong to the sets G0 and G1 then q s

e

Ai (X1 j , ..., Xsej ) ηij measβ˜j = 0,

(4.1.8)

j=1 i=1

where η1j , ..., ηsj are the direction cosines of ej , ( oriented from the common end point ) and β˜j is a cross-section of a cylinder with the axis ej by the hyperplane orthogonal to this axis, normalized by a factor µ−(s−1) ;

169


Figure 4.1.6.Equilibrium junction condition if this end point belongs to the set G1 then q s j=1 i=1

e

Ai (X1 j , ..., Xsej ) ηij measβ˜j = − tds−1 . 0

(4.1.9)

Here t is a constant vector equal to µs−1 T /measΣ1 ; 3) The vector-valued function defined on each segment e ⊂ B by u0 (x) =

s

Xie xi + X0e

i=1

(for x ∈ e ) is a continuous function of (x1 , ..., xs ) ∈ B , equal to zero for all nodes from G0 . Let this system (4.1.7)-(4.1.9) have a unique solution. Remark 4.1.3. The problem (4.1.7)-(4.1.9) does not depend on the form of cross-sections of the cylinders but it depends on their measure. Denote u ˜0 the extension of u0 (x) onto Bµ according to definition 4.1.4, and ˜0 the constant extension of the value of the function u0 (x0 ) for each denote U node x0 onto the connected part of the nodal domain, containing x0 . Let χ(x/µ) be a function equal to zero in Bµ \Bµ+ and to 1 in the domain Bµ++ , and let |∇z χ(z)| < c0 ,

0 ≤ χ ≤ 1,

χ ∈ C 1,

where c0 is a positive constant, independent of µ. Denote ˜0 (1 − χ(x/µ)). u(1) ˜0 χ(x/µ) + U µ (x) = u

(4.1.10)

170


Theorem 4.1.1. Let uµ be an exact solution of the problem (4.1.1) (1) (4.1.4) and let uµ be an asymptotic solution satisfying the assumptions (4.1.7) - (4.1.9). Then the estimate holds : √ uµ ; u(1) µ F L(Bµ ) ≤ C µ,

(µ → 0) (1)

where the constant C does not depend on µ, i.e. the pair (uµ ; uµ ) FL-converges √ and the rate of the FL-convergence is O( µ). Proof For each x from the nodal domain one obtains: (1)

|

∂u ˜0 ∂χ ∂uµ ˜0 )| ≤ c1 , (˜ u0 − U | = | χ(x/µ) + µ−1 ∂ξi ∂xi ∂xi

(4.1.11)

where c1 does not depend on µ. Estimate the integral I =

Bµ

s

(Ai (∇u(1) µ ) , ∂ϕ/∂xi ) dx.

i=1

Considering the inequality (4.1.11) we obtain I =

s

++ Bµ

(Ai (∇u(1) µ ) , ∂ϕ/∂xi ) dx + δ1 ,

i=1

where √ |δ1 | = O( µ)ϕH 1 (Bµ )

meas(Bµ )

As ϕ = 0 on S0 we obtain integrating by parts: I = − = −

x0

++ ∂Bµ

s

(Ai (∇u(1) µ ) , ϕ)ni ds =

i=1

++ ∂ Πx0 ∩∂Bµ

s

(Ai (∇u(1) µ ) , ϕ)ni ds + δ2 ,

i=1

where the summation is made with respect to all nodal points x0 , which do not belong to G0 , and Πx0 is the connected component of the set Bµ \Bµ++ which contains x0 , and n is an outside normal of the domain Πx0 (and so, n is an inside normal of Bµ++ .) The error δ2 is the same integral for nodes x0 ∈ G0 and it could be estimated as

|δ2 | ≤ C2 ϕL1 (∂Π

++ x0 ∩∂Bµ )

≤ C3 ϕL2 (∂Π

++ x0 ∩∂Bµ )

meas(Bµ ).

Now applying Lemma 4.A1.1 from the Appendix 4.A1 we obtain

171


|δ2 | ≤ C3 ϕL2 (∂Π

++ x0 ∩∂Bµ )

√ ≤ C4 µ∇ϕL2 (Πx0 )

where C2 , C3 , C4 do not depend on µ. For each node x0 we set < ϕ >Πx0

=

meas(Bµ ) ≤

meas(Bµ ),

1 meas(Πx0 )

ϕdx.

Πx0

We have the presentation ϕ(x) = < ϕ >Πx0 + (ϕ− < ϕ >Πx0 ) Lemma 4.A1.2 from the Appendix 4.A1 implies ϕ− < ϕ >Πx0 L2 (∂Π

++ x0 ∩∂Bµ )

√ ≤ C7 µ∇ϕL2 (Πx0 ) ,

where C7 does not depend on µ. Present the integral I in a form I = − −

x0

++ ∂ Πx0 ∩∂Bµ

s

++ ∂ Πx0 ∩∂Bµ

x0

s

(Ai (∇u(1) µ ) , < ϕ >Πx0 )ni ds −

i=1

(Ai (∇u(1) µ ) , (ϕ− < ϕ >Πx0 ))ni ds.

i=1

Here the first integral is equal to zero for all nodes which do not belong to G1 ∪ G2 (since the properties 1) and 2) of the vectors Xi ) and the first integral is equal to −µs−1 (t , < ϕ >Πx0 )d¯s−1 for x0 ∈ G1 , i.e. 0 −

x0

++ ∂ Πx0 ∩∂Bµ

s

=

(Ai (∇u(1) µ ) , < ϕ >Πx0 )ni ds =

i=1

Σ1

(t , ϕ)ds −

Σ1

Σ1

(t , < ϕ >Πx0 )ds =

(t , ϕ− < ϕ >Πx0 )ds.

Estimating x0

and

++ ∂ Πx0 ∩∂Bµ

s

(Ai (∇u(1) µ ) , (ϕ− < ϕ >Πx0 ))ni ds

i=1

Σ1

(t , ϕ− < ϕ >Πx0 )ds

172


by the sum C8 (

x0 ∈ / G1

+

meas(Bµ )ϕ − < ϕ >Πx0 L2 (∂Π

x0 ∈G1

++ x0 ∩∂Bµ )

+

meas(Bµ )ϕ − < ϕ >Πx0 L2 (S1 ∩∂Πx0 ) )

we canapply Lemma 4.A1.2 of the Appendix 4.A1 to obtain the estimate √ O( µ) meas(Bµ )ϕH 1 (Bµ ) . Thus for

I0 =

Bµ

we obtain I − I0 =

s

(Ai (∇uµ ) , ∂ϕ/∂xi ) dx =

(t , ϕ)ds

Σ1

i=1

Bµ

s i=1

(Ai (∇u(1) µ ) − Ai (∇uµ ) , ∂ϕ/∂xi ) dx =

√ = O( µ) meas(Bµ )ϕH 1 (Bµ )

and therefore the theorem is proved. Remark 4.1.4. It could be proved using the Poincar´ é - Friedrichs inequality for rod structures (Appendix 4.2), that L-convergence is a corollary of FLconvergence in the case of a scalar linear elliptic problem. s Indeed, let Ai (y) be j=1 Aij yj where (Aij )1≤i,j≤s is a constant symmetric positive definite matrix. Then using the Poincar´ é- Friedrichs inequality we estimate (taking ϕ = u1 − u2 ): u1 ; u2 F L(Bµ ) ≥ C9

≥ C10

∇(u1 − u2 )2L2 (Bµ ) ≥ u1 − u2 H 1 (Bµ ) meas(Bµ )

u1 − u2 H 1 (Bµ ) ≥ meas(Bµ )

≥ C11

u1 − u2 L2 (Bµ ) meas(Bµ )

where the constants C9 , C10 , C11 do not depend on µ. According to Theorem 4.1.1 √ u1 ; u2 F L(Bµ ) ≤ C µ.

Therefore

and thus

(1)

uµ − uµ L2 (Bµ ) √ ≤ C µ/C11 , meas(Bµ )

4.2. SHAPE OPTIMIZATION OF A FINITE ROD STRUCTURE.

173

uµ − u ˜0 L2 (Bµ ) √ = O( µ), meas(Bµ )

√ i.e. uµ L-converges to u0 and the rate of the FL-convergence is O( µ). The situation changes in the case of elasticity system (see [148],[44],[67]) where the Korn inequality constant depends on µ, nevertheless the discrete model (4.1.7)-(4.1.9) seems to be valid in the case when all rods of the structure work in the tension-compression regime. Thus Theorem 4.1.1 justifies in some sense asymptotic approximation (4.1.10). In case of bending or torsion regime of some rods more general discrete model should be applied. This model was obtained in [148] and (for the junction of two rods) in [168] (see also sections 4.4 and 2.3.5). Remark 4.1.5. Formulation (4.1.1)-(4.1.4) can be generalized: we can consider right-hand side in the equation (4.1.1) that has a support localized in some c0 µ-vicinities of the nodes and that its magnitude is of order 1/µ. Its average appears in the right-hand side of equations (4.1.8). This generalization has been developed in the Ph.D. thesis of R.Chiheb.

4.2

Shape optimization of a finite rod structure.

Below a new algorithm of optimal design based on the asymptotic analysis of rod structures and L-convergence is proposed. It could be applied in case when the given measure of the optimal domain is much less than the square in 2D case or cube in 3D case of the characteristic size of the problem. This case is too often: the modelling computations for various problems of optimal design by existing algorithms show ( [3], [4], [34], [186]) that as a rule the optimal structure is a set of connected bars, i.e. rod structure. To obtain this answer these algorithms start from the full 2D or 3D domain, square or cube for example. They iterate then (on the micro-level) the distribution of small anisotropic holes to obtain the optimal domain. Even if the optimal domain is not a rod structure normally the penalization procedure is applied to the optimal domain in order to transform the answer into some sort of a rod structure. The proposed below algorithm starts from the lattice structure from the very beginning and deals with lattice structure at each step. This principle simplifies the optimization procedure. On the other hand the proposed approach differs from that of the algorithms of optimal topologies of discrete structures (see review [70] ) because it starts from the continuous media formulation of the problem containing small parameter, and then obtains the discrete formulation as a result of the asymptotic analysis of the initial problem. This approach allows to estimate the accuracy of thus obtained discrete models and hence of the optimization result of the initial model. We present as well some numerical experiments. The structure of the section is as follows. We consider the boundary value problem for non linear partial derivative equations stated on a finite rod structure as in previous section and state the

174


problem of minimization of the stored energy (compliance) of the body with constraint of the given measure of a finite rod structure. The initial stored energy integral is replaced by the stored energy corresponding to the asymptotic solution obtained in the previous section. Then we introduce the initial configuration [70] as such a skeleton of finite rod structure that contains all admissible skeletons of finite rod structures. We minimize the approximated stored energy by an iterative algorithm of ”frozen fluxes”; it is a discrete analog of the iterative algorithm from [35]. At each step of this algorithm we fix the fluxes obtained in the previous iteration for all rods and redistribute then the thicknesses of the rods to minimize the stored energy . We give some results of numerical experiments (optimal dam, optimal cantilever arm, optimal bridge) calculated by the software OPTIFOR that implements the described method.

4.2.1

Stored energy as the cost

We consider the shape design problem of minimizing the stored energy (compliance) of finite rod structures of section 4.1 under the constraints of given vector-valued constant T and of given total measure of a finite rod structure that is µs−1 . The stored energy of the body E Bµ =

Bµ

s

(Ai (∇uµ ) , ∂uµ /∂xi ) dx =

(t , uµ ) dx.

(4.2.1)

Σ1

i=1

is a cost. We suppose the existence of the integral (4.2.1). The optimization problem seeks among all finite rod structures Bµ ∈ Q with a fixed measure µs−1 such one Bµ0 which minimizes the energy (4.2.1): Eopt = minBµ

∈ Q EB µ .

(4.2.2)

Thus the physical sense of the small parameter µ is the power (s − 1)−1 of the ratio of given measure of Bµ (given quantity of the material) to the power s of the characteristic size of the problem. We approximate the energy integral (4.2.1) by the stored energy integral (1) related to the asymptotic solution uµ

Bµ

s

(1) (Ai (∇u(1) µ ) , ∂uµ /∂xi ) dx.

i=1

In the linear scalar case under the assumptions of Remark 4.1.4 the difference √ of these two integrals is O( µµs−1 ). Indeed,

|

Bµ

s i=1

(Ai (∇uµ ) , ∂uµ /∂xi ) dx −

Bµ

s i=1

(1) (Ai (∇u(1) µ ) , ∂uµ /∂xi ) dx | ≤


≤ | +|

Bµ

Bµ

s i=1

s i=1

175

(1) ((Ai (∇uµ ) − Ai (∇u(1) µ )) , ∂uµ /∂xi ) dx | +

(Ai (∇uµ ) , (∂uµ /∂xi − ∂u(1) µ /∂xi )) dx | ≤

√ ≤ const ( µ meas(Bµ )u(1) µ H 1 (Bµ ) +

+ uµ H 1 (Bµ ) uµ − u(1) µ H 1 (Bµ ) ) =

√ √ = O( µmeas(Bµ )) = O( µµs−1 ). Taking into consideration the estimate (4.1.11) we obtain

Bµ

=

=

++ Bµ

= µs−1

(1) (Ai (∇u(1) µ ) , ∂uµ /∂xi ) dx =

i=1

s

(1) s (Ai (∇u(1) µ ) , ∂uµ /∂xi ) dx + O(µ ) =

i=1

eα h ⊂B

s

˜α B h,j

eα h ⊂B

s

(Ai (X1 , ..., Xs ) , Xi ) dx + O(µs ) =

i=1

|eα βj h |measβ

s

(Ai (X1 , ..., Xs ) , Xi ) + O(µs ),

i=1

|eα h|

where is the length of eα h. Enumerate all segments eα h

⊂B : e1 , ..., eN and denote by µs−1 m1 , ..., µs−1 mN the measures of the cross˜ α , i.e. µs−1 measβ βj and by (X0k , ..., Xsk ) sections of corresponding cylinders B h,j,µ ej ej the collection (X0 , ..., Xs ) for the segment ek . So the energy integral related (1) to uµ is approximated by

Bµ

where

s

(1) s−1 (Ai (∇u(1) EB + O(µs ), µ ) , ∂uµ /∂xi ) dx = µ

i=1

EB =

N

k=1

|ek |mk

s

(Ai (X1k , ..., Xsk ) , Xik ),

(4.2.3)

i=1

and |ek | stand for the lengths of segments ek . The approximated constraints are N

k=1

|ek |mk = 1,

(4.2.4)

176


as well as conditions 1)-3) of section 4.1 (i.e. (4.1.7)-(4.1.9)). Making the change Mk = |ek |mk , rewrite (4.2.3),(4.2.4) as EB =

N

k=1

and

Mk

s

(Ai (X1k , ..., Xsk ) , Xik ).

(4.2.3′ )

i=1

N

Mk = 1.

(4.2.4′ )

k=1

So, we replace the initial shape optimization problem (4.2.2) by the problem of minimization of cost (4.2.3’) under constraints (4.2.4’), (4.1.7)-(4.1.9).

4.2.2

Simplification of the set of finite rod structures. Initial configuration

Thus the initial partial derivative equation problem (4.1.1)-(4.1.4) is asymptotically reduced to the algebraic system (4.1.7)-(4.1.9) for a set of unknowns (X0k , ..., Xsk ), k = 1, ..., N. In the linear case when the matrix of this system is non-singular the problem is stable to the small perturbations of the skeleton B. Indeed the following proposition holds. Proposition 4.2.1. Let Bµ (ε) be a finite rod structure, such that each α segment eα h (ε) of its skeleton is obtained from eh by means of orthogonal transformation with the matrix Id + O(ε), i.e. identity matrix Id perturbed by a matrix that is O(ε), by translation on the distance O(ε), and contraction in ˜ α , and 1 + O(ε) times. We suppose that the cross-sections of the cylinders B h,j,µ α ˜ their perturbed images B (ε), are of the same measure. Let the perturbed imh,j,µ ages of the segments with the end points in G0 and G1 have also the end points in these domains and no other segments of the perturbed skeleton have the end points in these domains. Then the solution of perturbed system (4.1.7)-(4.1.9) (X0k (ε), ..., Xsk (ε)), k = 1, ..., N, tends to the solution of non-perturbed system (X0k , ..., Xsk ), k = 1, ..., N, as ε tends to zero. It is the trivial corollary of the regular perturbations theory of linear operators. This obvious result gives us an opportunity of approximation of the finite rod structure’s geometry. We introduce the initial configuration BI as such skeleton of finite rod structure, that contains all admissible skeletons of finite rod structures, i.e. we consider only such structures that their skeletons belong to the initial configuration. We suppose also that the nodes of the admissible structures belong to the set of nodes of the initial configuration. For example we can choose an ε−periodic lattice-structure of the next Chapter with µ 0.

179

(16)

k=1 i=1

Then the problem of minimization of (4.2.5) with constrains (4.2.6) has a unique solution calculated by the formula s ˆ k , ..., X ˆk ) , X ˆ k ))1/(1+α) (Ai (X 01 0s 0i ˆ k = ( i=1 . (4.2.8) M s N k k ˆ , ..., X ˆ ), X ˆ k ))1/(1+α) ( (A ( X i 01 0s 0i k=1 i=1

Proof. Using Lagrange multiplier λ > 0 we obtain the minimization problem of the Lagrange function n ˆ ˆ N ) + λ( EB (M1 , ..., M

N

k=1

ˆ k − 1). M

We seek the stationary point where its gradient vanishes and taking in account the relation (4.2.5) in a standard way we get the distribution (4.2.8) of ˆ k for the n-th step. The proposition is proved. the control parameters M

Thus we obtain the finite rod structure of the n-th step. ˆ k are less than the established positive value we If some of the values of M can remove such segments ek from the skeleton (if it does not lead to a lost of the connectivity). Remark 4.2.1. The asymptotic expansion of the solution of elasticity problem stated in the finite rod system considered in the next section could be used for the correction of the first order discrete model (4.1.7 ) - (4.1.9 ). Remark 4.2.2. The optimal design problem (4.1.1)-(4.2.4), (4.2.1),(4.2.2) can be generalized: we can consider equation (4.1.1) with the given right hand side function fµ (x), ”concentrated” in the neighborhoods of some fixed interior nodes (end points) O1 , ..., ON , included in all admissible finite rod structures. Let Ti , i = 1, ..., N be vector-valued constants. Let r0 be a positive value such that all balls with the centers in the nodes and with the radii µr0 belong to Bµ . Denote Y (Oi ) such ball with the center Oi . Define fµ (x) =

Ti µs−1 measY (Oi )

in each ball Y (Oi ), and define fµ (x) = 0 out of the union of these balls. Consider the problem (4.1.1’), (4.2.2) - (4.2.4) where (4.1.1’) is the non-homogeneous equation (4.1.1), i.e. s i=1

∂/∂xi ( Ai (∇x uµ )) = fµ (x),

f or x ∈ Bµ .

(4.1.1′ )

180


Then the analogous algorithm can be applied with the replacement of the ¯ relation (8) by the relation for nodes O q s

e Ai (X1 j , ..., Xsej ) ηij measβ˜j = TO¯ ,

(4.1.8′ )

j=1 i=1

¯ does not belong to the set {O1 , ..., ON }, and where TO¯ = 0 if the node O ¯ TO¯ = Tr if O = Or . The conclusion of Theorem 4.1.1 is valid for this modified problem. To prove it we present each integral (ffµ , ϕ) dx, x0 ∈ {O1 , ..., ON }, x0 = Oi , Πx0

in a form

Πx0

(ffµ , < ϕ >Πx0 ) dx +

= (T Ti , µs−1 < ϕ >Πx0 ) +

Πx0

Πx0

(ffµ , ϕ− < ϕ >Πx0 ) dx = (ffµ , ϕ− < ϕ >Πx0 ) dx

and estimate the last integral as ffµ L2 (Πx0 ) ϕ − < ϕ >Πx0 L2 (Πx0 ) ≤ ≤ O(µ)

Ti µs−1 meas(Bµ ) ∇ϕL2 (Πx0 ) = measY (Oi ) = O(µ) meas(Bµ )ϕH 1 (Πx0 ) .

These modifications complete the proof.

4.2.4

Some results of numerical experiment

Here we present some two-dimensional and three-dimensional numerical examples calculated by the OPTIFOR software which implements the described method. We give below the value of compliance f T x for every optimal structure. 1. The first example deals with the problem of the reinforcement of a dam. The initial configuration and the distribution of boundary forces are shown in Fig.4.2.2


181

Figure 4.2.2. Reinforcement of a dam: the initial configuration and the optimal structure; f T x = 12.22 2. The second example is concerned with the three-dimensional model of a cantilever arm. The left end is fixed while the vertical force is applied to the right end. The optimal result was obtained after 30 iterations.

Figure 4.2.3. Cantilever arm: the initial configuration and the optimal structure; f T x = 6.02

182


3. Figure 4.2.4 presents the optimal structure calculated for another initial configuration. The compliance of the optimal structure is here less than for the previous initial configuration. the difference of compliances is about 10 per cent.

Figure 4.2.4. Cantilever arm; another initial configuration: the initial configuration and the optimal structure; f T x = 5.39 Figure 4.2.5 gives the compliance as a function of the number of iterations. The same optimal result was obtained by a different method in [186].

Figure 4.2.5. Compliance histories.

4.3.

183

CONDUCTIVITY: AN ASYMPTOTIC EXPANSION

3. The third example is an ”optimal bridge”. The initial configuration is a square lattice. The structure is supported at the edges of its base. A vertical point force is applied in the middle of the lower side. The initial configuration and the boundary conditions are symmetric with respect to the axis of the force. Therefore the optimal structure is also symmetric, and the computations can be performed on half the initial configuration. The results are presented in the Fig. 4.2.6.

Figure 4.2.6. Initial configuration for a bridge with 480 initial bars and optimal structure of a bridge, f T x = 0.846

4.3

Conductivity: an asymptotic expansion

4.3.1

Construction of asymptotic expansion

In this section we consider finite rod structures Bµ defined in Definition 4.1.2 without the additional cubes. ˜ = α(x − h), whose first For any segment e = eα h , we define local variables x component is x ˜1 . Consider the conductivity equation in Bµ 5 f e (˜ x1 ), x ∈ S0 , (4.3.1) Lx u = ∆u = 0, x∈ / S0 , u|∂1 Bµ = 0,

(4.3.2)

∂u |∂ B = 0, ∂n 2 µ

(4.3.3)

184


where ∂1 Bµ is a part of the boundary that coincides with the base of one (or ˜ α (we assume that the nodal point corresponding to several) of the cylinders B hj this base is an end-point only for one segment e ⊂ B), ∂2 Bµ = ∂Bµ \∂1 Bµ . We call ”interior nodes” all nodes not belonging to ∂1 Bµ . We assume that f e (t) ∈ C ∞ . In what follows we omit the superscript e for f. We seek the solution u(x) in the class of functions H 1 (Bµ ). Below we construct the asymptotic expansion of the solution to problem (4.3.1)-(4.3.3) as µ → 0. Theorem 4.3.1. There exists a unique solution to problem (4.3.1)-(4.3.3). This is a well known result on existence and uniqueness of solution to a mixed boundary value conductivity problem ( [85],[86]). The domain Bµ here depends on the small parameter µ but as we see in Appendix A4.2, the Poincar´ é-Friedrichs inequality constant is independent of µ. This implies Theorem 4.3.2. The a priori estimate holds for a solution to problem (4.3.1)-(4.3.3): u2H 1 (Bµ ) ≤ Cf 2L2 (Bµ ) , where constant C does not depend on µ. We construct the asymptotic solution in two stages. First we construct an asymptotic solution outside neighborhoods of the nodes, and then we build the boundary layers. At the first step for every segment e an asymptotic solution is sought in the form of Bakhvalov’s ansatz (section 2.2); in the isotropic case all Nl = 0 except of N0 = 1 and so this ansatz is just a function ve of the longitudinal variable x˜1 , where x ˜ = αe (x − h). This function ve satisfies the second order differential equation (see section 2.2). For every rod this function is multiplied by a cutting function ρ˜ that is, equal to 1 on the main part of segment e and that vanishes in some const µ−vicinities of the ends. Of course, this multiplication perturbs the right hand side and generates some finite support discrepancy in the equation near the nodes. At the second step we first construct the Taylor expansion of the right hand side in the const µ−vicinities of the ends and we obtain the elliptic boundary layer problems with finite right hand sides set in the bundles of half-infinite cylinders related to every node. Such boundary layer problems have solutions stabilizing to constants on every outlet of the bundle if and only if the right hand side is orthogonal to constants. This orthogonality condition generates the flux balance equation (4.3.9) in every node. On the other hand we would like to have the boundary layer solutions exponentially decaying at infinity. It means that the stabilization constants should be equal to zero. It can be seen that these limit constants may be manipulated and changed by varying the values of functions ve in the ends of segments e. It gives us a possibility to vanish all stabilization limit constants by equating all values of functions ve in the node (probably with some constant right hand sides δve , see (4.3.10)). Finally we get the second order ordinary differential equations on the skeleton supplied with the junction conditions for the values of ve′ and ve in the

4.3.


185

nodes, as well as with the boundary conditions in the nodes of ∂1 Bµ and ∂2 Bµ . At the third step we seek an asymptotic solution to this problem set on the skeleton in the form of a regular series ansatz (4.3.12) and we obtain the all terms of the asymptotic expansion, and in particular, the leading term vl0 satisfying (4.3.23)-(4.3.27). The justification as usual (see section 2.4.4) is developed as follows: we truncate the ansatz at some level J neglecting the terms of order O(µJ ) (in H 1 −norm) and calculate all discrepancies in the right hand sides of the equation and eventually boundary conditions. We check that these discrepancies are of order O(µJ−1 ) in L2 −norm. Then we apply the a priori estimate and obtain that u − u(J) H 1 (Bµ ) = O(µJ−1 ). Finally we improve this estimate comparing u(J) to u(J−1) and estimating the difference u(J) − u(J−1) H 1 (Bµ ) = O(µJ−1 ). Hence, the estimate holds: u − u(J−1) H 1 (Bµ ) = O(µJ−1 ) and so, u − u(J) H 1 (Bµ ) = O(µJ ). 1. We seek a formal asymptotic solution of problem (4.3.1),(4.3.3) outside neighborhoods of the nodes on any section Se corresponding to the segment e = eα h of the set B, of the form u ˜(∞) ∼ e

∞ l=0

(µ)l Nle

x ˜′ dl v( ˜ ) 1

µ

d˜ xl1

,

(4.3.4)

e ˜′ where x ˜ = α(x − h) are the local co-ordinates introduced in eα h , Nl (ξ ) are the functions constructed in Chapter 2; let us denote ξ = (ξ1 , ξ ′ ), and ξ ′ = (ξ2 , . . . , ξs ), v is a smooth function of the variable x ˜1 . All these functions Nle e are equal to zero except of N0 = 1. Thus simply u ˜(∞) (x) = ve (˜ x1 ), x ∈ e, e

and it is a solution of equation ve ”(˜ x1 ) = f (˜ x1 ), x ∈ e. 2. Now construct the boundary layers in neighborhoods of the nodes and we (∞) extend u ˜e (x) to Bµ . We expand now the constructed above at the first stage (∞) u ˜e in a Taylor series in the neighborhood of any node x0 . We obtain

186


u ˜(∞) ∼ e

∞ r=0

µr

x1 (x0 )) (˜1 )r dr ve (˜ , d˜ xr1 r!

(4.3.5)

where ˜ = αe z, z = (x − x0 )/µ, Here αe stands for the matrix α of passage from the local (corresponding to the segment e) base to the global one. Denote by Πx0 the connected component of the set Bµ \Sˆ0 containing the point x0 . Denote by Πx0 ,z,0 the image of Πx0 under the transformation z = ˜ α be a cylinder whose intersection with Πx is nonempty. (x − x0 )/µ,. Let B 0 hj We extend it to a semi-infinite cylinder behind the basis which has no common α this extended cylinder, and by Πx0 ∞ the points with Πx0 . We denote by Bhj∞ α and Πx0 . union of all such cylinders Bhj∞ Moreover, we denote by Πx0 ,z,∞ the image of Πx0 ,∞ under the transformation z = (x − x0 )/µ.

4.3.


187

α Figure 4.3.1. Sets Πx0 , Bhj∞ , Πx0 ,∞ , Πx0 ,z,∞ α Let χ ˜e (z) be the characteristic function of Bhj∞ corresponding to the segment e of B. In the neighborhood of the node x0 we seek the solution in the form (∞)

uΠx ∼ 0

∞ l=0

µl Nl0 (z) +

χ ˜e (z)ˆ ρ(αe z)ve (˜ x1 (x0 )).

(4.3.6)

e(x0 )

Here the summation of the last sum in (4.3.6) extends over all segments e in B having x0 as an end-point, and ρˆ(t) = ρˆ(t1 ) (i.e. it depends only on the first

188


component of the variable t in IRs ) is an infinitely differentiable function that vanishes for |t1 | ≤ c0 , is equal to unity for |t1 | ≥ c0 + 1, and it satisfies the estimate 0 ≤ ρˆ(t) ≤ 1. The functions Nl0 exponentially stabilize to zero as |z| → ∞. Substituting (4.3.6) in (4.3.1), (4.3.3), and representing f as the sum of ρˆ(αe z)f and (1 − ρˆ(αe z))f, we find the recurrent chain of problems determining functions Nl0 : Lz Nl0 =

e(x0 )

χ ¯e (z)(Lz ((1 − ρˆ(αe z))

−(1 − ρˆ(αe z))Lz (

+δ¯l−2

x1 (x0 )) z˜1l dl ve (˜ ) d˜ x1r l!

x1 (x0 )) (˜1 )l dl ve (˜ )) l! d˜ xl1

dl−2 (˜1 )r (1 − ρˆ(αe z)) l−2 f˜(x0 ), z ∈ Πx0 ,z,∞ , r! d˜ x1

(4.3.71 )

(˜1 )l ∂ ∂N Nl0 ) ((1 − ρˆ(αe z)) = χ ¯e (z)(Lz l! ∂νz ∂νz e(x0 )

−(1 − ρˆ(αe z))

∂ (˜1 )l dl e x1 (x0 )), z ∈ ∂Πx0 ,z,∞ , )) l v˜ (˜ ( ∂νz l! d˜ x1

(4.3.72 )

where δ¯l = 1 for l ≥ 0 and δ¯l = 0 for l < 0. ∂ . Denote respectively Fleq (z) and Flb (z) the Here Lz = ∆z , ∂ν∂ z = ∂n z right-hand side functions of equation (4.3.71 ) and boundary condition (4.3.72 ). Remark that these functions vanish outside of the ball of radius c0 . Lemma 4.3.1.Problem (4.3.71 ), (4.3.72 ) is solvable in the class of functions that stabilize exponentially to a constant as |z| tends to infinity (in the sense of [88]-[90] ) if and only if Fleq (z)dz = Flb (z)ds. (4.3.8) Πx0 ,z,∞

∂ Πx0 ,z,∞

This condition implies (in case when x0 ∈ / ∂1 Bµ ) that µ χ ¯e (z)(1 − ρˆ(αe z))Lx ve dz e(x0 )

−

Πx0 ,z,∞

e(x0 )

∼

∂ Πx0 ,z,0 \Πx0 ,z,∞

e(x0 )

µ

Πx0 ,z,∞

χ ¯e (z)(1 − ρˆ(αe z))

∂ ve dz ∂νx

χ ¯e (z)(1 − ρˆ(αe z))f dz.

Here ∼ stands for the asymptotic equivalence up to terms of order O(µK )∀K. Taking into account the fact that

4.3.


189

Lx ve ∼ ve ” ∼ f, ∂ ve ∼ ve′ , ∂νx

we obtain:

e(x0 )

meas βe ve′ ∼ 0,

(4.3.9)

where βe stands for the cross-section of the cylinder corresponding to e (see the definition of a finite rod structure). If (4.3.9) is satisfied asymptotically exactly, then problems (4.3.71 ), (4.3.72 ) have solutions that stabilize exponentially to constants on every half-infinite cylinder as |z| → ∞. We keep the notation Ce for these constants x1 (x0 )) From the representations (4.3.5),(4.3.6) it follows that if the value of ve (˜ ∞ changes by a value denoted δve , then the value of Ce in the sum l=0 µl Nl0 (z) changes also by δve . Then we are left with the problem of finding δve such that (ve1 (˜ x1 (x0 )) + δve1 ) − (ve2 (˜ x1 (x0 )) + δve2 ) ∼ 0,

∀e1 (x0 ), e2 (x0 ), (4.3.10)

and the redefined in such manner solutions Nl0 stabilize to zero as |z| → ∞.

Figure 4.3.2. Segments e1 (x0 ), e2 (x0 ) If the node x0 is an end-point for only one segment e and x0 ∈ / ∂1 Bµ , then condition (4.3.10) is cancelled (the constant can be omitted from the solution). If x0 ∈ ∂1 Bµ , then on the part of ∂Πx0 ,z,∞ corresponding to ∂1 Bµ , under change of variable z = (x − x0 )/µ we prescribe the condition Nl0 = 0 instead of boundary condition (4.3.72 ) on this part of the boundary. Then the analogue of Lemma 4.3.1 holds for this problem, but the solvability condition (4.3.8)is no

190


long necessary. The solution stabilizes to a constant Ce . So the condition at the node x0 takes the form (ve (˜ x1 (x0 )) + δve ) ∼ 0, (4.3.11) where δve is chosen so that the solution Nl0 , modified in such a way that it satisfies boundary condition Nl0 = −Ce instead of Nl0 = 0 on the part of the boundary corresponding to ∂1 Bµ , stabilizes to zero. More precisely, we find first the limit constants atinfinity Cel for each Nl0 ∞ and then prescribe condition (4.3.10) with δve = − l=0 µl Cel . Seeking ve in the form of a power series in µ, ve =

∞

µl vel ,

(4.3.12)

l=0

we find that , as in Chapter 2, all the vel are determined from the recursive sequence of equations of the form vel ” = fe (˜ x1 )δl0 ,

x ∈ e,

(4.3.12)

supplied by the matching conditions at the nodes x0 that are end-points for at least two segments (and x0 ∈ / ∂1 Bµ )

measβe (vel )′ (˜ x1 (x0 )) = 0,

(4.3.13)

e(x0 )

and (vel 1 (˜ x1 (x0 )) + δvel 1 ) = (vel 2 (˜ x1 (x0 )) + δvel 2 ),

∀e1 (x0 ), e2 (x0 ),

(4.3.14)

and boundary conditions at the nodes x0 in ∂1 Bµ or in ∂2 Bµ , but such that they are end-points for only one segment, of the form (vel )′ (˜ x1 (x0 )) = 0,

x0 ∈ ∂2 Bµ

(4.3.15)

and (vel (˜ x1 (x0 )) + δvel ) = 0,

x0 ∈ ∂1 Bµ ,

(4.5.16)

where −δvel are found as the limit constants at infinity for Nl0 . Indeed, for any l, we solve first problem (4.3.7) for Nl0 (taking for vel the matching condition vel 1 (˜ x1 (x0 )) = vel 2 (˜ x1 (x0 )), ∀e1 (x0 ), e2 (x0 ), instead of (4.3.14)) and we obtain different limit constants Cel on the outlets (the subscript e means that the outlet corresponds to the segment e of skeleton B); we redefine then Nl0 (z) replacing vel in (4.3.14) by vel +δvel with δvel = −Cel ; this replacement leads to stabilization of redefined Nl0 (z) to zero. So after such redefinition we have Nl0 tending to zero on the outlets and we obtain condition (4.3.14) for vel . Extend δvel on B in such a way that (δvel )′ = 0 in some neighborhood of each node. Then problem (4.3.12)-(4.3.16) can be reduced to a problem for a sum we = vel + δvel :

4.3.

191


equation we ” = ψ(˜ x1 ),

x ∈ e,

(4.3.17)

supplied by the matching conditions at the nodes x0 that are end-points for at least two segments (and x0 ∈ / ∂1 Bµ )

measβe we′ (˜ x1 (x0 )) = 0,

(4.3.18)

e(x0 )

and x1 (x0 )) = we2 (˜ x1 (x0 )), we1 (˜

∀e1 (x0 ), e2 (x0 ),

(4.3.19)

and boundary conditions at the nodes x0 in ∂1 Bµ or in ∂2 Bµ , but such that they are end-points for only one segment, of the form we′ (˜ x1 (x0 )) = 0,

x0 ∈ ∂2 Bµ

(4.3.20)

we (˜ x1 (x0 )) = 0,

x0 ∈ ∂1 Bµ .

(4.3.21)

and Let H01 (B) be a space of functions defined on B , having a derivative on each segment e and continuous on B and vanishing at all nodes of ∂1 Bµ . Suppose that this space is supplied by the inner product (w, v)B = that is

e⊂B

e

measβe (w(˜ x1 )v(˜ x1 ) + w′ (˜ x1 )v ′ (˜ x1 ))d˜ x1 ,

B

measβe (w ˜e (˜ xe1 )˜ ve (˜ xe1 ) + w ˜e′ (˜ xe1 )˜ ve′ (˜ xe1 ))d˜ xe1 ,

where w ˜e (˜ xe1 ) = w(αe (x − h)) for x ∈ e. Then the variational formulation for problem (4.3.17)-(4.3.21) is as follows: to find w ∈ H01 (B) such that, for any v ∈ H01 (B) −

e∈B

′

′

w (˜ x1 )v (˜ x1 )measβe d˜ x1 =

e

e∈B

ψ(˜ x1 )v(˜ x1 )measβe d˜ x1 .

e

The existence and uniqueness of such solution is proved by a standard technique of Riesz theorem on representation of a bounded linear functional (or by Lax-Milgramm lemma), using the Poincar´ é- Friedrichs inequality on B v 2 (˜ x1 )d˜ x1 ≤ C(B) (v ′ (˜ x1 ))2 d˜ x1 , B

B

where the constant C(B) depends on B only. (This inequality is an immediate consequence of the Newton-Leibnitz formula for B: let e1 , . . . , en be segments

192


such, that the initial node of ei is the final node for ei−1 and the beginning of e1 is a node of ∂1 Bµ ; let x ∈ ej , then v(x) =

j−1

k=1

ek

∂˜ vek (l) (l)dl + ∂l

e

x ˜1j (x)

0

∂˜ vej (l) dl). ∂l

We denote by u(J) the truncated (partial) sum of the asymptotic series (4.3.5), obtained by neglecting the terms of order O(µJ ) (in the H 1 (Bµ )−norm. Choosing J sufficiently large, we can show for any K that equation (4.3.1) and conditions (4.3.2) and (4.3.3) are satisfied with remainders of order O(µK ) (in the L2 (Bµ )−norm). Hence , using the a priori estimate for problem (4.3.1)(4.3.3) given by Theorem 4.3.2, we obtain as in section 2.2, the following assertion. Theorem 4.3.3There holds the estimate u − u(J) H 1 (Bµ ) = O(µJ ).

(4.3.22)

This estimate justifies the asymptotic construction.

4.3.2

The leading term of the asymptotic expansion

Thus the limit problem for the leading term v0 of the asymptotic expansion is as follows: the equation ve 0 ” = f (˜ x1 ), x ∈ e, (4.3.23) supplied by the matching conditions at the nodes x0 that are end-points for at / ∂1 Bµ ) least two segments (and x0 ∈ ve′ 0 measβe (˜ x1 (x0 )) = 0, (4.3.24) e(x0 )

and ve1 0 (˜ x1 (x0 )) = ve2 0 (˜ x1 (x0 )),

∀e1 (x0 ), e2 (x0 ),

(4.3.25)

and boundary conditions at the nodes x0 in ∂1 Bµ or in ∂2 Bµ , but such that they are end-points for only one segment, of the form ve′ 0 (˜ x1 (x0 )) = 0,

x0 ∈ ∂2 Bµ

(4.3.26)

ve 0 (˜ x1 (x0 )) = 0,

x0 ∈ ∂1 Bµ .

(4.3.27)

and Extend v0 on Bµ in a following way. For any connected component of S0 , corresponding to the segment e we define v(x) = ve 0 (˜ x1 (x)), and for any connected component Πx0 of Bµ \S0 containing x0 we define v(x) = ve 0 (˜ x1 (x0 )), as a constant. Then the estimate (4.3.22) yields √ u − vL2 (Bµ ) / meas Bµ = O( µ).

(4.3.28)

4.3.


193

Remark 4.3.1. The right-hand side function can have more general structure on S0 :

x ˜′ ψ(˜ x1 ), (4.3.29) F µ

where ψ ∈ C ∞, F ∈ C 1. In this case F is decomposed as F (ξ˜′ ) = F¯ + F˜ (ξ˜′ ), F¯ =< F >β . Solution to problem (4.3.1)-(4.3.3) with right-hand side (4.3.29) on a cylinder corresponding to a segment e is sought in a form v˜e (˜ x1 ) +

∞

µl Ml

l=0

x ˜′ dl ψ( ˜ ) 1

µ

d˜ xl1

,

where Ml are solutions to problems ∆ξ˜′ M0 = F˜ (ξ˜′ ), ξ˜′ ∈ β, ∂M M0 = 0, ξ˜′ ∈ ∂β; ∂nξ˜ ˜′ ∆ξ˜′ Ml = −M Ml−2 + hM l , ξ ∈ β, l ≥ 2, ∂M Ml = 0, ξ˜′ ∈ ∂β; ∂nξ˜

where hM l =< Ml−2 >β . Then for v we have equation d2 ve (˜ x1 ) − F¯ ψ + d˜ x21

∞

l=2,l even

µl hM l

dl ψ( ˜1 ) = 0. d˜ xl1 (∞)

At the second stage we construct boundary layers uΠx in neighborhoods of 0 nodes x0

∞ ∞ x ˜′ dl ψ( ˜1 ) (∞) l 0 e l+2 . u Πx ∼ µ Nl (z) + χ ¯l (z)ˆ ρ(αe z) v (˜ x1 ) + µ Ml 0 µ d˜ xl1 l=0 l=0 e(x0 )

Applying the same procedure as above we obtain the homogenized problem of the same type as (4.3.12) but with some supplement in the right-hand side. The leading term is the same as in problem (4.3.1)-(4.3.3) with f = F¯ ψ.

194


4.4

Elasticity: an asymptotic expansion

4.4.1

Construction of asymptotic expansion

Let as above ξ ′ be (s−1)−dimensional vector of the last s−1 components of the vector ξ = (ξ1 , . . . , ξs ), i.e. ξ ′ = (ξ2 , . . . , ξs ). Denote by Sα the transformation defined by relations S2 ξ ′ = (−ξ2 , ξ3 ), S3 ξ ′ = (ξ2 , −ξ3 ) for s = 3, and by S2 ξ ′ = −ξ2 for s = 2. In what follows we assume that Sα β = β and Sα βj = βj , and we define Sα ξ = (ξ1 , Sα ξ ′ ), α = 2, ..., s − 1. This property means certain symmetry of the cross-sections of the rods of the finite rod structure Bµ . As in section 4.3, for any segment e = eα ˜ = h , we define local variables x α(x − h), whose first component is x ˜1 . We consider the system of equations of elasticity theory (elasticity equations)in Bµ :

s ∂uµ ∂ Aij Lx uµ = ∂xj ∂xi i,j=1

=

5

αe∗ (f˜1e (˜ x1 ), µ2 f˜2e (˜ x1 ), . . . , µ2 f˜3e (˜ x1 ))∗ , 0, uµ |∂1 Bµ = 0,

x ∈ S0 , x∈ / S0 ,

(4.4.1) (4.4.2)

∂uµ |∂ B = 0, (4.4.3) ∂ν 2 µ where ∂1 Bµ is a part of the boundary that coincides with the base of one (or ˜ α (we assume that the nodal point corresponding to several) of the cylinders B hj this base is an end-point only for one segment e ⊂ B), ∂2 Bµ = ∂Bµ \∂1 Bµ . We call ”interior nodes” all nodes not belonging to ∂1 Bµ . Here s ∂uµ ∂uµ ni , = Aij ∂xj ∂ν i,j=1

where Aij are constant s × s matrices, the elements akl ij of the Aij satisfy the relations akl ij = (δij δkl + δil δjk )M + λδik δjl , where M and λ are positive constants, that are the Lame´ coefficients, (n1 , . . . , ns ) is the normal to ∂Bµε , f e (t) ∈ C ∞ . We seek the solution uµ (x) in the class of s−dimensional vector-valued functions of H 1 (Bµ ). Below we construct the asymptotic expansion of the solution to problem (4.4.1)-(4.4.3) as µ → 0 under some assumptions P F1 and P F2 of geometrical rigidity of the frame. These assumptions will be formulated below for the limit problem. Theorem 4.4.1. There exists a unique solution to problem (4.4.1)-(4.4.3). This is a well known result on existence and uniqueness of solution to a mixed boundary value problem of elasticity ( [55]).

4.4.

ELASTICITY: AN ASYMPTOTIC EXPANSION

195

The domain Bµ here depends on the small parameter µ and therefore an a priori estimate can also depend on this small parameter. On the other hand as we have seen in section 4.2 the a priori estimate for analogous problem for conductivity equation does not depend on the small parameter because the Poincare-Friedrichs ´ inequality constant is independent of µ. Is elasticity equation different in this sense? To answer this question we should study the Korn inequality constant for the domain Bµ . This constant depends on µ. Indeed, the following assertion holds. Theorem 4.4.2. For any µ small enough for any finite rod structure Bµ there is a vector valued function vµ ∈ H 1 (Bµ ) such that vµ |∂1 Bµ = 0, ∇vµ 2L2 (Bµ ) ≥ c1 µs−1 and EBµ (vµ ) ≤ c2 µs that is EBµ (vµ ) ≤ (

c2 )µ∇vµ 2L2 (Bµ ) , c1

where c1 and c2 are positive constants independent of µ. Here EBµ (φ) =

s

Bµ i,j=1

(eji (φ))2 dx, eji (φ) =

∂φj 1 ∂φi ). + ( ∂xi 2 ∂xj

Proof. α Without lost of generality we can prove this assertion for one rod Bh,j,µ such that one of its bases belongs to ∂1 Bµ (we pass to the general case extending vµ α by zero all over Bµ \Bh,j,µ ). α We can consider the cylinder Bj,µ instead of Bh,j,µ because the orthogonal transformation and the translation do not change the order of the values ∇vµ 2L2 (Bµ ) and EBµ (vµ ).

Figure 4.4.1. Domain π.

196


So, let us consider cylinder π = (−1, 1) × βµ , where βµ = {(x′ /µ ∈ β}, β is (s − 1)−dimensional bounded domain with a piecewise-smooth boundary (satisfying the cone condition; in the two-dimensional case β is an interval), i.e. there exist constants c1 , c2 , c3 > 0, independent of µ, and the s-dimensional vector-valued function wµ (x), such that wµ ∈ H 1 (π) and wµ |x1 =±1 = 0, wµ 2L2 (π) ≥ c1 µs−1 ,

∇wµ 2L2 (π) ≥ c2 µs−1 ,

Eπ (wµ ) ≤ c3 µs . For example , for s = 2, µ < 1/2 one can build the following function, satisfying these conditions: ⎧

⎪ ρ(x /µ)ρ((1 + x )/µ)ρ((x − 1)/µ) −x ⎪ 2 1 1 1 ⎪ for x1 ≤ 0 ⎪ ⎨ 1 + x1

wµ = ⎪ ⎪ ⎪ x2 ρ(x1 /µ)ρ((1 + x1 )/µ)ρ((x1 − 1)/µ) for x1 > 0, ⎪ ⎩ 1−x 1

where ρ(t) is an even continuously differentiable function, 0 ≤ ρ(t) ≤ 1, equal to zero for |t| ≤ 1/3 and equal to 1 for |t| ≥ 2/3. (In the case s = 3 the two first components are the same and the third one is equal to zero.) Indeed, ∂w2 2 2 ) dx = 2 meas β µs−1 , ∇wµ L2 (π) ≥ ( π ∂x1 wµ 2L2 (π) ≥ w22 dx = meas β µs−1 , π

and

Eπ (wµ ) = Eπ∩({x1 ≤−1+µ}∪{|x1 |≤µ}∪{x1 ≥1−µ}) (wµ ) = O(µs ), ∂w2 ∂w2 ∂w1 1 because ∂w ∂x2 = 0, and because the ∂x1 = ±1, ∂x2 = O(1), ∂x1 = O(1), s strains differ from zero on the set of the measure O(µ ). By the linear application α , and in we could map this constructed vector-function to any cylinder Bh,j,µ particular, the cylinder having a base in ∂1 Bµ . Thus, the theorem is proved. This theorem shows that the Korn inequality constant depends on µ. Nevertheless the Korn inequality holds true in this case with the constant that is some negative power of µ. Theorem 4.4.3. Any vector-valued function u ∈ H 1 (Bµ ), such that u|∂1 Bµ = 0, satisfies the Korn inequality

∇u2L2 (Bµ ) ≤ CK µ−q EBµ (u), where CK and q are independent of µ.

4.4.


197

Proof. The proof of this theorem was given first by constructing an extension of u from Bµ to some domain which does not depend on µ (see [145]). We give it here below in Appendix 5.A2. This theorem with the Poincar´ é-Friedrichs inequality implies Theorem 4.4.4. The a priori estimate holds for a solution to problem (4.4.1)-(4.4.3): u2H 1 (Bµ ) ≤ Cµ−q f 2L2 (Bµ ) , where constant C does not depend on µ. Let us remind the definition at section 2.3 of the matrix-valued functions of rigid displacements of an infinite bar { ξ = (ξ1 , . . . , ξs ) ∈ IRs , ξ1 ∈ IR; ξ ′ = (ξ2 , . . . , ξs ) ∈ β }. For s = 2 it is the identity matrix 1 0 Φ = I = , 0 1 while for s = 3 it depends on the rapid ⎛ 1 Φ(ξ) = ⎝ 0 0

−1/2 Here a =

β

(ξ22 +ξ32 )dξ2 dξ3 meas β

variable ξ and it is equal to ⎞ 0 0 0 1 0 −ξ3 a⎠ . 0 1 ξ2 a

.

Denote d the number of columns of Φ (i.e. d = 2 for s = 2 and d = 4 for s = 3). Moreover, we define as well the extended matrix of rigid displacements in IRs : for s = 2 it is 1 0 −ξ2 J˜(ξ) = , 0 1 ξ1 while for s = 3 it is ⎛

1 0 0 0 J˜(ξ) = ⎝ 0 1 0 −ξ3 0 0 1 ξ2

−ξ2 ξ1 0

⎞ −ξ3 0 ⎠. ξ1

˜ of Chapter 2 only by a normalization factor a of the third for (it differs from Φ s = 2 or the fourth for s = 3 column. ) Now we formulate the two assumptions on Bµ . As in the previous section we will consider segments e which belong to the set of segments eα h constituting the skeleton B. Let us define for any segment e of B the matrix Γe of dimension d × d such that for any z ∈ IRs , J˜∗ (z)αe∗ = Γe J˜∗ (αe z), where z is a column (z1 , . . . , zs )∗ . It can be proved immediately that

198


Lemma 4.4.1. ⎛

Γe = ⎝

for s = 2, and

Γe =

0

⎞ 0 0⎠ 1

αe∗ 0

αe∗ O

O ¯e Γ

¯ e is an orthogonal 3 × 3 matrix and O the 3 × 3 zero matrix. for s = 3, where Γ Proof. For s = 2, let cos θ sin θ αe = , θ ∈ IR; −sin θ cos θ let us check the last scalar equality of the relation J˜∗ (z)αe∗ = Γe J˜∗ (αe z). The last row of J˜∗ (z)αe∗ is (−z2 , z1 )αe∗ ∗ ∗ −z2 cos θ + z1 sin θ −(αe z)2 = = z2 sin θ + z1 cos θ (αe z)1 because αe z =

z1 cos θ + z2 sin θ , −z1 sin θ + z2 cos θ

and therefore the components (αe z)1 and (αe z)2 of this vector satisfy the above relation. The first two scalar equalities of relation J˜∗ (z)αe∗ = Γe J˜∗ (αe z) are evident because Iαe∗ = αe∗ I. So, for s = 2 lemma is proved. If s = 3 let us check firstly the last three scalar equalities of the relation J˜∗ (z)αe∗ = Γe J˜∗ (αe z). For any k ∈ {1, 2, 3}, denote Jk (z) = Πk z, where ⎛

Π1 = ⎝

0 0 0

⎛ ⎞ 0 0 0 −1 0 −1⎠ , Π2 = ⎝ 1 0 0 0 1 0

⎞ ⎛ 0 0 0⎠ , Π3 = ⎝ 0 0 1

⎞ 0 −1 0 0 ⎠. 0 0

Then the matrix J˜(z) contains the last three columns Jk (z), satisfying the following relations: αe∗ Jk (αe z) = αe∗ Πk αe z =

3 i=1

γki (αe )Πi z,

4.4.


199

where γki (αe ) are the components of the 3 × 3 matrix ⎛

α32 α23 − α22 α33 γ(αe ) = ⎝ α22 α13 − α12 α23 α32 α13 − α12 α33

α31 α22 − α21 α32 α21 α12 − α11 α22 α31 α12 − α11 α32

and αij are the components of the matrix αe . One can see that

⎞ α31 α23 − α21 α33 α21 α13 − α11 α23 ⎠ , α31 α13 − α11 α33

γij (αe ) = γji (αe∗ ), i, j ∈ {1, 2, 3}, so

γ ∗ (αe ) = γ(αe∗ ). On the other hand, the above relation αe∗ Jk (αe z) =

can be rewritten in the form Jk (αe z) =

3 i=1

3

γki (αe )Πi z

i=1

γki (αe )αe Ji (z),

and therefore Jk (z) which is equal to Jk (αe∗ αe z) can be presented in the form 3 ∗ ∗ i=1 γki (αe )αe Ji (αe z), that is 3

i,j=1

γki (αe∗ )γij (αe )αe∗ αe Jj (z) =

3

i,j=1

γki (αe∗ )γij (αe )J Jj (z).

These vector-valued functions Jj (z) are linearly independent functions. Therefore, the product γ(αe∗ )γ(αe ) is the identity matrix I. On the other hand we have just proved that γ ∗ (αe ) = γ(αe∗ ). So, γ ∗ (αe )γ(αe ) = I, and the matrix γ(αe ) is an orthogonal matrix. Consider now the 3×3−matrix J˜last (z) constituted of the three last columns ∗ of the matrix J˜(z). Its transposition J˜last (αe z) consists of three lines: ⎛ ⎞ J1∗ (αe z) ∗ ⎝ J2∗ (αe z)⎠ = γ(αe )(αe J˜last (z))∗ = γ(αe )J˜last (z)αe∗ ; ∗ J3 (αe z) so,

∗ ∗ γ ∗ (αe )J˜last (αe z) = J˜last (z)αe∗ ,

¯ e = γ ∗ (αe ). and therefore, Γ

200

CHAPTER 4. FINITE ROD STRUCTURES The first three scalar equalities of relation J˜∗ (z)αe∗ = Γe J˜∗ (αe z)

are evident because Iαe∗ = αe∗ I. So, for s = 3 lemma is proved as well. Assume now that the two following conditions hold true. ˜ x1 ) be any s-dimensional vector-valued function Condition P F1 . Let Ψ(˜ defined on B vanishing in the nodes belonging to ∂1 Bµ ; - which has generalized derivative along e for every e ⊂ B, - which is such that all its components, except the first, are linear on every segment e, ˜ e1 = - which satisfies at all interior nodes x0 the matching conditions αe∗1 Ψ ∗ ˜ e2 αe2 Ψ for any two segments e1 (x0 ) and e2 (x0 ), having the common end-point x0 . Then it is assumed that the following inequality holds: ˜ e1 (˜ ˜ e1 /d˜ (Ψ x1 ))2 d˜ x1 ≤ c (dΨ x1 (˜ x1 ))2 d˜ x1 e

e

e

e

˜ e is the first component of the with a constant c depending only on B. Here Ψ 1 e ˜ . vector-valued function Ψ Thus, Condition P F1 is satisfied for example, if only one node does not belong to ∂1 Bµ . Suppose that for s = 3 the following Condition P F2 , analogous to the Poincar´ é- Friedrichs inequality, holds. ˜ x1 ) be any three-dimensional 1-periodic vectorCondition P F2 . Let Ψ(˜ valued function defined on B ˜ 1 has generalized derivative along e for every - whose first component Ψ segment e ⊂ B, ˜ k , k = 2, 3, have two generalized - whose second and third components Ψ derivatives along e for every segment e ⊂ B, - whose second and third components vanish at all nodes, and - which satisfies at all interior nodes x0 the matching conditions of the form ⎛ ⎞ ⎛ ⎞ ˜ e1 ˜ e2 Ψ Ψ 1 1 ¯ ∗e ⎝ dΨ ¯ ∗e ⎝ dΨ ˜ e1 /d˜ ˜ e2 /d˜ Γ x1 ⎠ = Γ x1 ⎠ 2 2 1 2 e 1 ˜ /d˜ ˜ e2 /d˜ dΨ x1 dΨ x1 3

3

for any two segments e1 (x0 ) and e2 (x0 ), having the common end-point x0 ; - which satisfies at all boundary nodes x0 ∈ ∂1 Bµ the boundary conditions of the form

˜ ˜ ˜ = 0; ∂ Ψ2 = 0; ∂ Ψ3 = 0. Ψ ∂x ˜1 ∂x ˜1 Thenit is assumed that the following inequality holds:

˜e ˜e dΨ dΨ 3 2 2 2 e 2 ˜ (˜ x1 )) + ( (˜ x1 )) d˜ x1 Ψ1 (˜ x1 )) + ( d˜ x1 d˜ x1 B

4.4.


201

˜e ˜e ˜e d2 Ψ d2 Ψ dΨ1 3 2 2 2 2 (˜ x1 )) d˜ x1 (˜ x1 )) + ( (˜ x1 )) + ( ≤ c d˜ x21 d˜ x1 d˜ x1 2 B with a constant c depending only on B. We suppose also for simplicity of presentation that for any interior node there are three non-coplanar (if s = 3), respectively two non-collinear (if s = 2), segments e having this node as an end-point. This condition will be called the non-coplanarity condition. Remark 4.4.1 In case s = 2 the following conditions analogous to P F2 ˜ e be any scalar function depending on the is satisfied automatically. Let Ψ variable x ˜1 , having two derivatives along segment e ⊂ B, satisfying the matching conditions at the interior nodes x0

˜ e2 (x0 ) ˜ e1 (x0 ) dΨ dΨ , = d˜ x1 d˜ x1

and vanishing at all nodes (the derivative vanishes as well at the nodes x0 ∈ ∂1 Bµ . ) Then the following inequality holds

(

B

≤ c

˜e dΨ (˜ x1 ))2 d˜ x1 d˜ x1

B

(

˜e d2 Ψ (˜ x1 ))2 d˜ x1 d˜ x21

with a constant c depending only on B. We construct the asymptotic solution as usual in two stages. First we construct an asymptotic solution outside neighborhoods of the nodes, and then we build the boundary layers. At the first step for every segment e an asymptotic solution is sought in the form of the ansatz of section 2.3 related to some d−dimensional (d = 2 if s = 2; d = 4 if s = 3) vector-valued function ωe depending on the longitudinal variable x˜1 , where x ˜ = αe (x − h). The components of vector-valued function ωe satisfy some ordinary differential equations (homogenized high order equations). For every segment e this ansatz is multiplied by a cutting function ρ˜ that is, equal to 1 on the main part of segment e and that vanishes in some const µ−vicinities of the ends. Of course, this multiplication perturbs the right hand side and generates some finite support discrepancy in the equation near the nodes. At the second step we first construct the Taylor expansion of the right hand side in the const µ−vicinities of the nodes and we obtain the elliptic boundary layer problems with finite right hand sides set in the bundles of half-infinite cylinders related to every node. Such boundary layer problems have solutions stabilizing to some rigid displacement on every outlet of the bundle if and only if the right hand side is orthogonal to all rigid displacements. This orthogonality condition generates the stress balance equation (4.4.11) in every inner node. On the other hand we would like to have the boundary layer solutions exponentially decaying at infinity. It means that the stabilization limit rigid displacements

202


should be equal to zero. It can be seen that these limit constants may be manipulated and changed by varying the values of functions ωe and its derivatives ∂ωe2 ∂ωe2 ˜1 in the ends of segments e. It gives us a possibility to vanish all sta∂x ˜1 , ..., ∂ x bilization limit rigid displacements by equating all values of functions ωe and its ∂ω 2 ∂ω 2 derivatives ∂ x˜1e , ..., ∂ x˜1e in the node multiplied by some passage matrix Γe . This matrix is defined as a matrix of orthogonal transform of rigid displacements. Finally we get the second order ordinary differential equations on the skeleton supplied with the junction conditions for the values and for some derivatives of ωe in the nodes, as well as with the boundary conditions in the nodes of ∂1 Bµ and ∂2 Bµ . At the third step we seek an asymptotic solution to this problem set on the skeleton in the form of a regular series ansatz and we obtain the all terms of the asymptotic expansion, and in particular, the leading term satisfying (4.4.55)(4.4.72). The justification as usual (see section 2.4.4) is developed as follows: we truncate the ansatz at some level J neglecting the terms of order O(µJ ) (in H 1 −norm) and calculate all discrepancies in the right hand sides of the equation and eventually boundary conditions. We check that these discrepancies are of order O(µJ−1 ) in L2 −norm. Then we apply the a priori estimate and obtain that

u − u(J) H 1 (Bµ ) = O(µJ−1−q ). Factor µ−q appeared here due to the dependency of the Korn inequality constant on µ. This dependency could be made more precise ( [75], [118] ), but for our study the exact value of q is not too important because we have constructed the complete asymptotic expansion, for any J satisfying the equation with accuracy O(µJ−1 ). Finally we improve this estimate comparing u(J+q+1) to u(J) and estimating the difference u(J+q+1) − u(J) H 1 (Bµ ) = O(µJ ). On the other hand, u(J+q+1) − uH 1 (Bµ ) = O(µJ ). Hence, the estimate holds: u − u(J) H 1 (Bµ ) = O(µJ ). 1. We seek a formal asymptotic solution of problem (4.4.1),(4.4.3) on any section Se corresponding to the segment e = eα h of the set B, of the form u ˜(∞) ∼ e

∞ l=0

(µ)l Nle

x ˜′ dl v˜( ˜ ) 1

µ

d˜ xl1

,

(4.4.4)

4.4.

203


where Nle (ξ˜′ ) are the s × d matrix valued functions in Chapter 2 and v˜(˜ x1 ) is a d−dimensional vector-valued function such that its components have the following expansions: v˜1 (˜ x1 ) ∼ v˜r (˜ x1 ) ∼

∞

∞

µq v1q (˜ x1 ),

q=0

µq−2 vrq (˜ x1 ), r = 2, . . . , s,

q=0

and in case s = 3 the fourth component is v˜4 (˜ x1 ) ∼

∞

µq−1 v1q (˜ x1 ).

(4.4.5)

q=0

Substituting (4.4.4) in (4.4.1) and (4.4.3) we find as in Chapter 2 that (4.4.3) is satisfied asymptotically exactly and that (4.4.1) takes form ∞

µl−2 hN l

l=2

dl v˜( ˜1 ) ˜e − f ∼ 0, d˜ xl1

(4.4.6)

˜ f e = f e for s = 2 and ˜ fe = where fe is a d−dimensional vector such that ˜ e f for s = 3. In what follows we omit the superscript e for f. 0 (∞)

2. We expand now the constructed above at the first stage u ˜e series in the neighborhood of any node x0 . We obtain u ˜(∞) ∼ e

∞

¯ e (˜) (µ)r N r

r=0

x1 (x0 )) dr v˜(˜ , d˜ xr1

in a Taylor

(4.4.7)

where ˜ = αe z, z = (x − x0 )/µ,

¯re (˜) = N (∞)

(∞)

r Nqe (˜′ ) (˜1 )q , (r − q)! q=0

˜e . and ue = αe∗ u Here αe stands for the matrix α of passage from the local (corresponding to the segment e) base to the global one. Denote by Πx0 the connected component of the set Bµ \S0 containing the point x0 . Denote by Πx0 ,z,0 the image of Πx0 under the transformation z = ˜ α be a cylinder whose intersection with Πx is nonempty. (x − x0 )/µ. Let B 0 hj We extend it to a semi-infinite cylinder behind the basis which has no common α points with Πx0 . We denote by Bhj∞ this extended cylinder, and by Πx0 ∞ the α union of all such cylinders Bhj∞ and Πx0 . Moreover, we denote by Πx0 ,z,∞ the image of Πx0 ,∞ under the transformation z = (x − x0 )/µ.

204


α Let χ ˜e (z) be the characteristic function of Bhj∞ corresponding to the segment e of B. In the neighborhood of the node x0 we seek the solution in the form (∞)

u Πx ∼ 0

∞

µl Nr0 (z) +

l=0

χ ˜e (z)ˆ ρ(αe z)u(∞) , e

(4.4.8)

e(x0 )

where the summation of the last sum in (4.4.8) extends over all segments e in B having x0 as an end-point, and ρˆ(t) = ρˆ(t1 ) (i.e. it depends only on the first component of the variable t in IRs ) is a differentiable function that vanishes for |t1 | ≤ c0 , is equal to unity for |t1 | ≥ c0 + 1, and such that 0 ≤ ρˆ(t) ≤ 1. Substituting (4.4.8) in (4.4.1), (4.4.3), we find the recurrent chain of problems determining functions Nl0 : Lz Nl0 =

l

l1 =0 e(x0 )

⎛

+αe∗ ⎝

¯le (z)) χ ¯e (z)(Lz ((1 − ρˆ(αe z))N 1

l1 ¯le (z)) d v˜e −(1 − ρˆ(αe z))Lz N 1 d˜ xl11

l−l1

(˜ x1 (x0 ))

l−2

z˜1 dl−2 ˜ f (x ) (1 − ρˆ(αe z)) d˜ δ¯l−2 (l−2)! xl−2 1 0 1

l−4

z˜1 dl−4 ˜ (f (x ), . . . , f˜s (x0 ))∗ δ¯l−4 (l−4)! (1 − ρˆ(αe z)) d˜ xl−4 2 0 1

⎞

⎠ , z ∈ Πx0 ,z,∞ ,

(4.4.91 )

l ∂N Nl0 ∂ ¯le (z)) ((1 − ρˆ(αe z))N = χ ¯e (z)(Lz 1 ∂νz ∂νz l1 =0 e(x0 )

−(1 − ρˆ(αe z))

d l1 ∂ ¯e Nl1 (z)) l1 v˜e ∂νz d˜ x1

l−l1

(˜ x1 (x0 )), z ∈ ∂Πx0 ,z,∞ ,

(4.4.92 )

where δ¯l = 1 for l ≥ 0 and δ¯l = 0 for l < 0. Denote respectively Fleq (z) and Flb (z) the right-hand side functions of equation (4.4.91 ) and boundary condition (4.4.92 ). Remark that these functions vanish outside of the ball of radius c0 . Let J be the linear span of columns of the matrix J˜. Lemma 4.4.2.Problem (4.4.91 ), (4.4.92 ) is solvable in the class of vectorvalued functions that stabilize exponentially as |z| tends to infinity (in the sense of [130], [129] ) to a vector-valued function in J if and only if J˜∗ (z)F Fleq (z)dz = J˜∗ (z)F Flb (z)ds. (4.4.10) Πx0 ,z,∞

∂ Πx0 ,z,∞

This stabilization (4.4.10) takes place so that Nl (z) − j(z) dz → 0, Πx0 ,z,+∞ ∩{σ≤z1 ≤σ+1}

4.4.

205


where j ∈ J . Thus condition (4.4.10) implies (in case when x0 ∈ / ∂1 Bµ ) that µ χ ¯e (z)(1 − ρˆ(αe z))J˜∗ (z)Lz u(∞) dz e Πx0 ,z,∞

e(x0 )

−

e(x0 )

∼

∂ Πx0 ,z,0 \Πx0 ,z,∞

µ

e(x0 )

Πx0 ,z,∞

χ ¯e (z)(1 − ρˆ(αe z))J˜∗ (z)

∂ (∞) u dz ∂νx e

f dz. χ ¯e (z)(1 − ρˆ(αe z))J˜∗ (z)αe∗˜

Taking into account the fact that ∼ α∗ Lx u(∞) e

∂ (∞) u ∼ α∗ ∂νx e

∞

l−1

µ

l=1

∞

µl−2 hN l

l=2

dl v˜( ˜1 ) , d˜ xl1

∂N Nl dl v˜( ˜1 ) , A1j + A11 Nl−1 ∂ξξj d˜ xl1 j=1

s

as well as the relation of Remark 2.3.2, we obtain

e(x0 )

Γe meas βe Λµ,e v˜e +

∞ j=1

µj

e(x0 )

ˆ µ,e,j (˜ Λ ve , ˜ f ) ∼ 0,

(4.4.11)

¯ is replaced by where Λµ,e is the operator in (2.3.71) where the constant M and ae is given by formula of section 2.3: (ξ 2 + ξ 2 ) dξ dξ −1/2 2 3 2 3 ; a= mes β β

¯ M ae ,

ˆ µ,e,j (˜ ve , ˜ f ) are some linear differential operators whose rows have the same and Λ order with respect to µ as those of Λµ,e . We remind that Γe is the d × d matrix of the transformation J˜∗ (z)αe∗ = Γe J˜∗ (αe z)

of the matrix J˜ and βe the cross-section of the cylinder corresponding to e (see the definition of a finite rod structure). Denote ˆ ve = Λ˜ Γe meas βe Λµ,e v˜e , e(x0 )

where µ = 1. If (4.4.11) is satisfied asymptotically exactly, then problems (4.4.8),(4.4.9) have solutions that stabilize exponentially to limit functions of J on every halfinfinite cylinder as |z| → ∞. We keep the notation J˜(z)Ce for these limit

206


¯ functions; here Ce is a d−dimensional vector corresponding to the segment e; we also consider the vectors

∗ ∂˜ v2e e e (4.4.12) Ve = v˜1 , v˜2 , µ ∂x ˜1

for s = 2, and Ve =

ve ∂˜ v e ∂˜ v˜1e , v˜2e , v˜3e , v˜4e , µ 2 , µ 3 ˜1 ∂x ˜1 ∂ x

∗

(4.4.13)

for s = 3. As in section 4.3, from the representations (4.4.8),(4.4.9) it follows ∞that if the value of Ve changes by δV Ve , then the value of J˜(z)Ce in the sum l=0 µl Nl0 (z) changes by αe∗ J˜(αe z)Ce δV Ve , that is , Ce changes by Γe δV Ve . Then we are left with the problem of finding δV Ve such that Ve1 + δV Ve1 ) − Γe2 (V Ve2 + δV Ve2 ) ∼ 0, Γe1 (V

∀e1 (x0 ), e2 (x0 ),

(4.4.14)

and the redefined in such manner solutions Nl0 stabilize to zero as |z| → ∞. If the node x0 is an end-point for only one segment e and x0 ∈ / ∂1 Bµ , then condition (4.4.14) is cancelled (the rigid displacement can be omitted from the solution). If x0 ∈ ∂1 Bµ , then on the part of ∂Πx0 ,z,∞ corresponding to ∂1 Bµ , under change of variable z = (x − x0 )/µ we prescribe the condition Nl0 = 0 instead of boundary condition (4.4.92 ) on this part of the boundary. Then the analogue of Lemma 4.4.2 holds for this problem, but the solvability condition (4.4.10)is no long necessary. The solution stabilizes to a rigid displacement J˜Ce so the condition at the node x0 takes the form Ve + δV Ve ) ∼ 0, Γe (V

(4.4.15)

where δV Ve is chosen so that the solution stabilizes to zero as in the previous section. Seeking a formal asymptotic solution of problem (4.4.6), (4.4.11), (4.4.14), (4.4.15) in the form (4.4.5) of a power series in µ, we find that , as in Chapter 2, all the vel are determined from the recursive sequence of equations of the form R˜ v e l = f˜l (˜ x1 ),

x ∈ e,

(4.4.16)

where R is an operator of (2.3.69) with µ = 1; these equations are supplied by the matching conditions at the nodes x0 that are end-points for at least two segments (and x0 ∈ / ∂1 Bµ ) ˆ v e l = Ψl (x0 ), Λ˜

(4.4.17)

and Γe1 (V Vel1 + δV Vel1 ) = Γe2 (V Vel2 + δV Vel2 ),

∀e1 (x0 ), e2 (x0 ),

(4.4.18)

4.4.

207


and boundary conditions at the nodes x0 in ∂1 Bµ or in ∂2 Bµ , but such that they are end-points for only one segment, of the form ˆ v e l = Ψl (x0 ), Λ˜ Vel

+

δV Vel

= 0,

x0 ∈ ∂1 Bµ

(4.4.19)

x0 ∈ ∂2 Bµ ,

(4.4.20)

where

Vel

=

⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩

∂v ˜e l+1 v˜1e l , v˜2e l+2 , ∂2x˜1

v˜1e l , v˜2e

l+2

, v˜3e

l+2

, v˜4e

∗

,

˜e l+1 ˜e l+1 ∂ v l+1 ∂ v , ∂2x˜1 , ∂3x˜1

s = 2,

∗

,

s = 3,

here f˜l and Ψl are defined in terms of the functions v˜e l1 with l1 < l, f˜0 = (f˜1 , 0, . . . , 0), and Ψ0 = 0. The constant vectors δV Vel are defined in terms of the 0 solution Nl of problems (4.4.9), found under the assumption that δV Vel = 0 in (4.4.18) and (4.4.20); afterwards Nl0 (z) are redefined so that they stabilize to zero as |z| → +∞ (as above). Indeed, for any l, we solve first problem (4.4.9) for Nl0 and we obtain different limit rigid displacements J˜Cel on the outlets (the subscript e means that the outlet corresponds to the segment e of B); we redefine then Nl0 (z) replacing Vel at the end of the segment e in (4.4.9) by Vel + δV Vel where δV Vel is such that l l Γe δV Ve = −Ce ; this replacement leads to stabilization of redefined Nl0 (z) to zero. So after such redefinition we have Nl0 tending to zero on the outlets and we obtain conditions (4.4.18) and (4.4.20) for Vel . Problems (4.4.16)-(4.4.20) split into a pair of problems, for v˜1l and for (˜ v2l , . . . , v˜dl )∗ : (i) for any l we solve first a problem for an s−dimensional vector w ˜ l (˜ x1 ) e l (denoted w ˜ (˜ x1 ) on each segment e ⊂ B) consisting of -the equation 2 e l ˜1 (˜ x1 ) ¯d w E = f˜l,1 (˜ x1 ), x ∈ e, d˜ x21

(4.4.21)

-the conditions ¯ E

meas βe γe

e(x0 )

dw ˜1e l (˜ x1 ) = Ψl (x0 ) d˜ x1

(4.4.22)

at all nodes except x0 ∈ ∂1 Bµ , where γe is a director vector of segment e with initial point x0 ; -the condition l−2 w ˜ e l = −(δV Vel1 , δV Vel−2 Ves ), 2 , ..., δV

at all nodes x0 ∈ ∂1 Bµ , and

(4.4.23)

208

CHAPTER 4. FINITE ROD STRUCTURES -the condition αe∗1 w ˜ e1 l = αe∗2 w ˜ e2 l − αe∗1 δV Vel1 + αe∗2 δV Vel2

(4.4.24)

at all nodes x0 which are common points for two different segments e1 (x0 ) and e2 (x0 ); here Ψl (x0 ) and δV Vel are given constants, f˜l,1 is a given right-hand side function; all components of the vector valued function w˜e l (except of w ˜1e l ) are linear functions; ˜ l (˜ x1 ) (ii) secondly we solve a problem for a (d − 1)−dimensional vector W e l ˜ (denoted W (˜ x1 ) on each segment e ⊂ B) -for s = 2 consisting of -the equation ¯ < ξ 2 >β −E 2

˜ e l (˜ d4 W x1 ) = f˜l−2,2 (˜ x1 ), x ∈ e, d˜ x41

(4.4.25)

˜ e l (˜ d2 W x1 ) = Ψl (x0 ) d˜ x21

(4.4.26)

-the matching condition ¯ E

meas βe < ξ22 >β

e(x0 )

at all nodes except x0 ∈ ∂1 Bµ , -the condition

˜ e2 l (˜ ˜ e1 l (˜ dW x1 ) dW x1 ) = + η l (x0 ), d˜ x1 d˜ x1

(4.4.27)

at all nodes x0 which are common points for two different segments e1 (x0 ) and e2 (x0 ); -the condition ˜ e2 l = (w ˜ e1 W

l−2

)ort

e2

+ ξ(x0 ),

(4.4.28)

for any e2 (x0 ) and for some e1 (x0 ) non-collinear to e2 (x0 ), here (w ˜ e1 l−2 )ort e2 is e1 l−2 the orthogonal projection of the vector w ˜ γe on the axis orthogonal to e2 ; η l (x0 ), ξ(x0 ) are given constants; for s = 3, this problem consists of the equations 2 ˜ e l x1 ) ¯ d W (˜ = 0, x ∈ e, M 2 d˜ x1

(4.4.29)

¯ < ξ 2 >β −E 2

˜ e l (˜ d4 W x1 ) 2 = f˜l−2,2 (˜ x1 ), x ∈ e, d˜ x41

(4.4.30)

¯ < ξ 2 >β −E 3

˜ e l (˜ d4 W x1 ) 3 = f˜l−2,3 (˜ x1 ), x ∈ e, d˜ x41

(4.4.31)

4.4.


-the matching conditions

¯ W ¯eΛ ˜ e l = Ψl (x0 ), Γ e

209

(4.4.32)

e(x0 )

˜ e2 l = (w W ˜ e1 2

l−2

)ort,2 + ξ2 (x0 ),

(4.4.33)

for any e2 (x0 ) and for some e1 (x0 ) such that its projection (e1 (x0 ))ort,2 on the local axis x ˜e22 is non-zero; ˜ e2 l = (w W ˜ e1 3

l−2

)ort,3 + ξ3 (x0 ),

(4.4.34)

for any e2 (x0 ) and for some e1 (x0 ) such that its projection (e1 (x0 ))ort,3 on the local axis x ˜e32 is non-zero;

˜ e1 l (˜ ˜ e1 l x1 ) dW x1 ) ∗ 3 ¯ e (W ˜ e1 l , dW2 (˜ ) , Γ 1 1 d˜ x1 d˜ x1

˜ e2 l (˜ ˜ e2 l x1 ) dW x1 ) ∗ 3 ¯ e (W ˜ e2 l , dW2 (˜ , ) + η(x0 ), ∀e1 (x0 ), e2 (x0 ), =Γ 2 1 d˜ x1 d˜ x1

(4.4.35)

at nodes x0 having at least two initial vectors e1 (x0 ), e2 (x0 ); here 2 ˜ e l ˜ e l (˜ ¯ ˜ e l x1 ) d2 W x1 ) x1 ) ¯ 3 ¯ W ¯ < ξ 2 >β d W2 (˜ ˜ = meas βe ( M dW1 (˜ ), , E < ξ32 >β ,E Λ e 2 2 d˜ x21 d˜ x1 ae d˜ x1 (4.4.36) -and the boundary conditions

˜e ˜e dW dW 3 2 ˜ e l (˜ = η3l (x0 ), x0 ∈ ∂1 Bµ ; (4.4.37) = η2l (x0 ), W x1 ) = ξ l (x0 ), 2 d˜ x1 d˜ x1

-and ¯ W ¯eΛ ˜ e l = Ψ0 (x0 ); Γ e

(4.4.38)

at all nodes having only one initial segment, x0 ∈ / ∂1 Bµ . Above the functions and constants in right-hand sides are defined by the previous terms of asymptotic expansion with scripts less than l. We compile then v˜el = w ˜ el + (0, wel )∗ for s = 2 and v˜el = (w ˜ el , 0)∗ + (0, wel )∗ for s = 3. These problems, in turn, could be reduced to the problems with homogeneous matching and boundary conditions (by subtracting of function satisfying to nonhomogeneous conditions from the unknown function), i.e. to 1) a problem for an s−dimensional vector ν˜e (˜ x1 ), consisting of the equation 2 e x1 ) ¯ d ν˜1 (˜ = g˜(˜ x1 ), x ∈ e, E d˜ x21

(4.4.39)

210

CHAPTER 4. FINITE ROD STRUCTURES the conditions ¯ E

meas βe γe

e(x0 )

d˜ ν e (˜ x1 ) = 0, d˜ x1

(4.4.40)

at all nodes except x0 ∈ ∂1 Bµ , -the condition ν˜ = 0

(4.4.41)

αe∗1 ν˜e1 = αe∗2 ν˜e2

(4.4.42)

at all nodes x0 ∈ ∂1 Bµ , and -the condition

at all nodes x0 which are common points for two different segments e1 (x0 ) and e2 (x0 ); all components of the vector valued function ν˜e (except of ν˜1e ) are linear functions on every e; 2) a problem for a (d−1)−dimensional vector ν˜e (˜ x1 ), which for s = 2 consists of the equations 4 e x1 ) ¯ < ξ 2 >β d ν˜ (˜ = g˜(˜ x1 ), x ∈ e, −E 2 d˜ x41

(4.4.43)


meas βe < ξ22 >β

e(x0 )

d2 ν˜e (˜ x1 ) = 0, ν˜ = 0, d˜ x21

(4.4.44)

at all nodes except x0 ∈ ∂1 Bµ , -the condition d˜ ν e1 (˜ x1 ) d˜ ν e2 (˜ x1 ) , = d˜ x1 d˜ x1

(4.4.45)

at all nodes x0 which are common points for two different segments e1 (x0 ) and e2 (x0 ); -the condition ν˜e = 0,

d˜ ν e (˜ x1 ) = 0, d˜ x1

(4.4.46)

at all nodes x0 ∈ ∂1 Bµ , and d2 ν˜e (˜ x1 ) ¯ = 0, Emeas βe < ξ22 >β d˜ x21

(4.4.47)

at all nodes having only one initial segment e(x0 ) except x0 ∈ ∂1 Bµ ; for s = 3, this problem consists of the equations 2 e x1 ) ¯ d ν˜1 (˜ = g˜1 (˜ x1 ), x ∈ e, M d˜ x21

(4.4.48)

4.4.


211

¯ < ξ 2 >β −E 2

d4 ν˜2e (˜ x1 ) = g˜2 (˜ x1 ), x ∈ e, d˜ x41

(4.4.49)

¯ < ξ32 >β −E

d4 ν˜3e (˜ x1 ) = g˜3 (˜ x1 ), x ∈ e, d˜ x41

(4.4.50)

-the matching conditions ¯ ν˜e = 0, ν˜e = 0, ν˜e = 0, ¯eΛ Γ e 2 3

(4.4.51)

e(x0 )

ν3e1 (˜ x1 ) ∗ ν2e1 (˜ x1 ) d˜ e1 d˜ ¯ e (˜ , ) Γ ν , 1 1 d˜ x1 d˜ x1 e2 ¯ e (˜ =Γ 2 ν1 ,

x1 ) ∗ ν e2 (˜ x1 ) d˜ d˜ ν2e2 (˜ , 3 ) , ∀e1 (x0 ), e2 (x0 ), d˜ x1 d˜ x1

(4.4.52)

at nodes x0 having at least two initial vectors e1 (x0 ), e2 (x0 ); and the boundary conditions ν˜e = 0,

d˜ ν3e (˜ x1 ) d˜ ν2e (˜ x1 ) = 0, = 0, d˜ x1 d˜ x1

(4.4.53)

at all nodes x0 ∈ ∂1 Bµ , and ¯ ν˜e = 0, ¯eΛ Γ e

(4.4.54)

at all nodes having only one initial segment, x0 ∈ / ∂1 Bµ . Let conditions P F1 and P F2 be satisfied. Denote H1 (B) and H1,2,2 (B) the spaces of functions described in conditions P F1 and P F2 respectively. Denote H2 (B) the space of functions described in Remark 4.4.1. Lemma 4.4.2 If conditions P F1 and P F2 hold then problem (4.4.39)-(4.4.42) and problem ( 4.5.43)-(4.4.47) if s = 2, (respectively (4.4.48)-(4.4.54) if s = 3) are solvable. Proof. Indeed, problem (4.4.39)-(4.4.42) is equivalent to the following variational formulation:

=

B

−

B

¯ meas βe E

d˜ ν1e (˜ x1 ) x1 ) dψ˜1e (˜ d˜ x1 d˜ x1 d˜ x1

meas βe g˜1 (˜ x1 )ψ˜1e (˜ x1 )d˜ x1 , ∀ψ˜1e ∈ H1 (B).

For s = 2, problem (4.4.43)-(4.4.47) is equivalent to the following variational formulation:

2 e x1 ) x1 ) d2 ψ˜e (˜ ¯ d ν˜ (˜ d˜ x1 meas βe < ξ22 >β E 2 2 d˜ x d˜ x B 1 1

212


=

B

meas βe g˜(˜ x1 )ψ˜e (˜ x1 )d˜ x1 , ∀ψ˜e ∈ H2 (B).

For s = 3 problem (4.4.48)-(4.4.54) is stated as −

B

=

¯ meas βe M

B

x1 ) d˜ ν1e (˜ x1 ) dψ˜1e (˜ d˜ x1 d˜ x1 d˜ x1

meas βe g˜1 (˜ x1 )ψ˜1e (˜ x1 )d˜ x1 ,

2 e x1 ) x1 ) d2 ψ˜2e (˜ ¯ d ν˜2 (˜ d˜ x1 meas βe < ξ22 >β E 2 2 d˜ x1 d˜ x1 B = meas βe g˜2 (˜ x1 )ψ˜e (˜ x1 )d˜ x1 ,

B

=

B

2 e x1 ) x1 ) d2 ψ˜3e (˜ ¯ d ν˜3 (˜ d˜ x1 meas βe < ξ32 >β E 2 2 d˜ x d˜ x B 1 1

meas βe g˜3 (˜ x1 )ψ˜e (˜ x1 )d˜ x1 , ∀(ψ˜1e , ψ˜2e , ψ˜3e ) ∈ H1,2,2 (B).

The unique solvability of these variational problems follows from the Riesz functional representation theorem for a bounded (due to P F1 and P F2 conditions) linear functional on a Hilbert space. We denote by u(J) the truncated (partial) sum of the asymptotic series (4.4.4),(4.4.5),(4.4.8) obtained by neglecting the terms of order O(µJ ) (in the H 1 (Bµ )−norm. Choosing J sufficiently large, we can show for any K that equation (4.4.1) and conditions (4.4.2) and (4.4.3) are satisfied with remainders of order O(µK ) (in the L2 (Bµ )−norm. Hence , using the a priori estimate for problem (4.4.1)-(4.4.3) given by Theorem 4.4.3, we obtain as in section 2.2, the following assertion. Theorem 4.4.5There holds the estimate u − u ˜(J) H 1 (Bµ ) = O(µJ ). More general cases are studied in [119] (non-symmetrical section of a bar, when the conditions of the very beginning of the present section are not respected) and in [121] (the two-dimensional case).

4.4.2

The leading term of asymptotic expansion

Thus the limit problem for the leading term v0 of the asymptotic expansion is as follows. e e For s = 2, we define a vector-valued function v0e = (v01 , v02 ) for every sege ment e; the first component v01 is a solution to the equation

4.4.


2 e x1 ) ¯ d v˜01 (˜ = f˜1e (˜ x1 ), x ∈ e, E d˜ x21

213

(4.4.55)

with the matching condition ¯ E

e d˜ v01 (˜ x1 ) e1 e2 = 0, v01 = v01 d˜ x1

meas βe γe

e(x0 )

(4.4.56)

at all nodes x0 initial at least for two segments e1 and e2 , except x0 ∈ ∂1 Bµ , and with the condition e v˜01 =0

(4.4.57)

at all nodes x0 ∈ ∂1 Bµ , and the condition e

d˜ v (˜ x1 ) ¯ = 0, Emeas βe γe 01 d˜ x1

(4.4.58)

for the nodes initial for the only one segment e; e v02 is a solution to the equation ¯ < ξ 2 >β −E 2

e d4 v˜02 (˜ x1 ) = f˜2e (˜ x1 ), x ∈ e, d˜ x14

(4.4.59)


meas βe < ξ22 >β

e(x0 )

e d2 v˜02 (˜ x1 ) = 0, d˜ x21

e1 d˜ v e2 (˜ x1 ) d˜ v02 (˜ x1 ) e = 0, , v˜02 = 02 d˜ x1 d˜ x1

(4.4.60)

at all nodes x0 which are common points for two different segments e1 (x0 ) and e2 (x0 ); -the condition e v˜02 = 0,

d˜ ν e (˜ x1 ) = 0, d˜ x1

(4.4.61)

at all nodes x0 ∈ ∂1 Bµ , and d2 ν˜e (˜ x1 ) e ¯ = 0, = 0, v˜02 Emeas βe < ξ22 >β d˜ x21

(4.4.62)

at all nodes having only one initial segment e(x0 ) except x0 ∈ ∂1 Bµ ; Extend v0e onto Bµ in a following way: for any connected component of S0 we define v(x) = αe∗ v0e (˜ x1 (x)), and for any connected component Πx0 of Bµ \S0 , x1 (x0 )), as a constant. Then the estimate containing x0 , we define v(x) = αe∗ v0e (˜ of Theorem 4.4.5 yields

214

CHAPTER 4. FINITE ROD STRUCTURES √ u − vL2 (Bµ ) / meas Bµ = O( µ).

e e e e For s = 3, we define a vector-valued function v0e = (v01 , v02 , v03 , v04 ), for e every segment e; the first component v01 is a solution to the equation 2 e x1 ) ¯ d v˜01 (˜ = f˜1e (˜ x1 ), x ∈ e, E d˜ x21

(4.4.63)

with the matching condition ¯ E

meas βe γe

e(x0 )

e d˜ v01 (˜ x1 ) = 0, d˜ x1

(4.4.64)

at all nodes x0 initial at least for two segments e1 and e2 , except x0 ∈ ∂1 Bµ , and with the condition v˜01 = 0

(4.4.64)

at all nodes x0 ∈ ∂1 Bµ , and the condition e

d˜ v (˜ x1 ) ¯ = 0, Emeas βe γe 01 d˜ x1

(4.4.65)

for the nodes initial for the only one segment e; e e e (v04 , v02 ) is a solution to the problem , v03 2 e x1 ) ¯ d v˜04 (˜ M = 0, x ∈ e, d˜ x21

(4.4.66)

¯ < ξ 2 >β −E 2

e d4 v˜02 (˜ x1 ) = f˜2e (˜ x1 ), x ∈ e, d˜ x41

(4.4.67)

¯ < ξ32 >β −E

e d4 v˜03 (˜ x1 ) = f˜3e (˜ x1 ), x ∈ e, d˜ x41

(4.4.68)

-the matching conditions ¯ (v e , v e , v e )∗ = 0, v˜e = 0, v˜e = 0, ¯eΛ Γ e 04 02 03 02 03

(4.4.69)

e(x0 )

at all interior nodes x0 , and -the matching conditions e1

e1

v (˜ x1 ) ∗ v (˜ x1 ) d˜ e1 d˜ ¯ e (˜ ) , 03 Γ v04 , 02 1 d˜ x1 d˜ x1 e2 ¯ e (˜ =Γ v04 , 2

e2 v e2 (˜ x1 ) ∗ d˜ v02 (˜ x1 ) d˜ , 03 ) , ∀e1 (x0 ), e2 (x0 ), d˜ x1 d˜ x1

at nodes x0 having at least two initial vectors e1 (x0 ), e2 (x0 ); and

(4.4.70)

4.5. FLOWS IN TUBE STRUCTURES

215

-the boundary conditions

e e v˜04 = 0, v˜02 = 0,

e e d˜ v03 (˜ x1 ) d˜ v02 (˜ x1 ) e = 0, = 0, v˜03 = 0, d˜ x1 d˜ x1

(4.4.71)

at all nodes x0 ∈ ∂1 Bµ , and e e ¯ (v e , v e , v e )∗ = 0, ¯eΛ = 0, Γ v˜02 = 0, v˜03 e 04 02 03

(4.4.72)

at all nodes having only one initial segment, x0 ∈ / ∂1 Bµ . Extend v0e onto Bµ in a following way: for any connected component of S0 we define e e e e v(x) = αe∗ {(v01 (˜ x1 (x)), v02 (˜ x1 (x)), v03 (˜ x1 (x)))∗ +µ−1 (0, −˜ x3 (x), x ˜2 (x))v04 (˜ x1 (x))},

and for any connected component Πx0 of Bµ \S0 , containing x0 , we define v(x) = αe∗ v0e (˜ x1 (x0 )), as a constant. Then the estimate of Theorem 4.5.5 yields √ u − vL2 (Bµ ) / meas Bµ = O( µ).

4.5

Flows in tube structures

In case of flows in thin domains like finite rod structures these domains are called here below tube structures. The Navier-Stokes problem stated in tube structures (or finite rod structures), i.e. in connected finite unions of the thin cylinders with the ratio of the diameter to the height of the order µ r0 . Suppose also that these functions (and r0 ) do not depend on µ. Thus we have defined f in small domains obtained from supp fj by a contraction in 1/µ times with the centers Oj . We define f as zero in all other points. Let f ∈ C 1 , ν > 0. Let Hdiv=0 (B µ ) be space of vector valued functions from H 1 (B µ ) with van0 µ µ ishing divergence, let Hdiv =0 (B ) be the subspace of Hdiv =0 (B ) of functions vanishing on the boundary. Suppose that g can be continued in B µ as a vector valued function gˆ of Hdiv=0 (B µ ). The variational formulation is as follows: find 0 µ u ∈ Hdiv=0 (B µ ) such that u − gˆ ∈ Hdiv =0 (B ), and such that it satisfies to the integral identity

−

ν

Bµ

s i=1

∂ϕ ∂u , ) dx + ( ∂xi ∂xi

s

B µ i=1

∂ϕ ) dx = ui (u , ∂xi

(f, ϕ) dx,

Bµ

0 µ for all ϕ ∈ Hdiv of the scalar product in IRs , ui =0 (B ). Here (, ) is the symbol s ∂ϕ is the i−th component of the vector u, i.e. i=1 ui (u , ∂x ) = (u , (u, ∇)ϕ). i The existence and uniqueness of the solution was proved in [87] (for sufficiently small values of µ) . We construct the asymptotic expansion in a form

ua =

K l=2

µl {

e=ej , j=1,...,n

uel (

n x − Oi xe,L i )}, uBLO ( )χµ (x) + l µ µ i=0

(4.5.6)

219


pa =

K+1

µl−2

pel (xe1 )χµ (x) +

e=ej , j=1,...,n

l=2

+

K+1 l=2

µl−1

n i=0

n

i pBLO ( l

i=0

pe2 (Oi )(1 − χµ (x))θi (x)

x − Oi ), µ

(4.5.7)

Here χµ is a function equal to zero at the distance not more than (dˆ0 + 1)µ from Oj , j = 0, 1, ..., n; dˆ0 µ = max {d0 µ, d1 µ}, d0 µ is the infimum of radiuses of all spheres with the center O such that every point of it belongs only to not more than one of the cylinders Bj j = 1, ..., n; d1 is the maximal diameter of the domains γ0 , γ1 , ..., γn , and θj (x) = 0 if |x − Oj | > mini |ei |/2, θj (x) = 1 if |x − Oj | ≤ mini |ei |/2.

e

e

We have introduced here the local system of coordinates Ox1j ...xsj associe ated with a segment ej such that the direction of the axis Ox1j coincides with ej the direction of the segment OOj , i.e. x1 is a longitudinal coordinate. The axes e e Ox1j , ..., Oxsj form a cartesian coordinate system. We suppose that the function e e χµ is equal to zero on the cylinder Bjµ if x1j ≤ (d0 +1)µ or if |x1j −|ej || ≤ (d0 +1)µ (here |ej | is the length of the segment ej ), we suppose that the function χµ is e e equal to one on this cylinder if x1j ≥ (d0 + 2)µ and |x1j − |ej || ≥ (d0 + 2)µ, and ej e we define χµ by relations χµ (x) = χj (x1 /µ) if (d0 + 1)µ ≤ x1j ≤ (d0 + 2)µ, and ej ej we pose χµ (x) = χj ((x1 − |ej |)/µ) if (d0 + 1)µ ≤ |ej | − x1 ≤ (d0 + 2)µ. Here χj is a differentiable on IR function of one variable, it is independent of µ, and it is equal to zero on the segment [−(d0 + 1), d0 + 1] and it is equal to one on the union of the intervals (−∞, −(d0 + 2)) ∪ ((d0 + 2), +∞). For any γjµ , χµ e e is equal to zero on it. The variable xej L = (x2j , ..., xsj ). Let the relation between the columns (here T is the transposition symbol) xT and xej ,T be xT = Γj xej ,T + O, j = 1, ..., n,

(4.5.8)

where Γi is an orthogonal matrix of passage from the canonic base to the local one (in the previous sections we used for matrices Γj the notation αe∗ ). Then the vector valued function uel and the scalar functions pel are defined up to the scalar constants cle , del : e

e

e

e

e

e

e

uej (ξ L ), 0, ..., 0)T , pl j (x1j ) = cl j x1j + dl j , l = 2, 3, ... . ul j (ξ L ) = cl j Γj (˜ (4.5.9), where ξ L = (ξ2 , ..., ξs ) and u ˜ej is the solution of the problem

220


ν∆ξL u ˜ej − 1 = 0, ξ L ∈ βj ,

u ˜ej |∂βj = 0, j = 1, ..., n,

(4.5.10)

e

and d2j = 0.

Figure 4.5.2. Dilated domains ΩOj . The boundary layer solution is a pair constituted of a vector valued function BLOj BLOj and scalar function pl satisfying to the Stokes system: ul 0 0 ν∆ξ uBLO − ∇ξ pBLO = f0 (ξ)δl,0 + l l

+

e=ej , j=1,...,n

e

e

{cl j {−ν∆ξ (χj (ξ1j )Γj (˜ uej (ξ L ), 0, ..., 0)T ) e

+(

e

+ ∇ξ (χj (ξ1j )ξ1j )} + e

cepj cerj )(χj (ξ1j )˜ uej (ξ L )

p+r=l−1

e

∂ ej uej (ξ L ), 0, ..., 0)T + e χj (ξ1 ))Γj (˜ ∂ξ1j e

j ∇ξ (χj (ξ1j ))}, + dl+1

(4.5.11)

221


0 divξ uBLO = − divξ { l

e

e=ej , j=1,...,n

e

{cl j χj (ξ1j )Γj (˜ uej (ξ L ), 0, ..., 0)T }, ξ ∈ ΩO0 (4.5.12)

with the Dirichlet condition 0 uBLO |∂ΩO0 = 0 l

(4.5.13)

and for i = 1, ..., n BLOj

ν∆ξûl

BLOj

− ∇ξˆpl

ˆ l,0 + = fj (ξ)δ

e

e

ˆ j (˜ + cˆl j {−ν∆ξˆ(χj (ξˆ1j )Γ uej (ξˆL ), 0, ..., 0)T ) e e + ∇ξˆ(χj (ξˆ1j )ξˆ1j )} +

+(

e


p+r=l−1

∂ ej uej (ξ L ), 0, ..., 0)T + e χj (ξ1 ))Γj (˜ ∂ξ1j

ej e + dˆl+1 ∇ξˆ(χj (ξˆ1j )),

BLOj

divξûl

e e ˆ = − divξˆ{ˆ cl j χj (ξˆ1j )Γ uej (ξˆL ), 0, ..., 0)T }, ξˆ ∈ ΩOj j (˜

(4.5.14)

(4.5.15)

with the Dirichlet condition BLOj

ul

|∂ΩO

BLOj

ul

e

j

,ξˆ1 j =0

|∂ΩO

e

j

= gj δl,2 ,

(4.5.16)

= 0,

(4.5.17)

,ξˆ1 j =0

˜ j ∪ γ0 , and Ω ˜ i are the half-infinite cylinders obtained from where ΩO0 = ∪nj=1 Ω µ Bj by infinite extension behind the base βˆjµ and by homothetic dilatation in 1/µ ˜ j by a symmetric times (with respect to the point O ); let Ωj be obtained from Ω µ reflection relatively the plain containing βj and let ΩOj = Ωj ∪ γjt , where γjt is obtained from γj by a translation (such that the point Oj becomes O). The e e variable ξˆ1j is opposite to ξ1j , i.e. to the first component of the vector ΓTj ξ ej ,T . e ˆ T (ξ ej )T , where Γ ˆ j = IdΓ ˆ j and Id ˆ is the diagonal matrix with the So ξˆ1j = Γ j diagonal elements −1, 1, ..., 1. The constants cêl , dêl are defined in such a way that e e e e e e e e e e the linear functions pl j (x1j ) = cl j x1j + dl j and pl j (x1j ) = cˆl j (|ej | − x1j ) + dˆl j are equal, i.e. e e e e e cl j , dˆl j = cl j |ej | + dl j . (4.5.18) cl j = −ˆ

222


We suppose that every term in the sum e=ej , j=1,...,n in (4.5.12) is defined only on the branch of ΩO0 , corresponding to e = ej , and it vanishes in γ0 . We seek the exponentially decaying at infinity solutions of these boundary e e layer problems and we choose the constants cel , cˆl j , del , dˆl j from the conditions ej of existence of such solutions [120]. We define first cˆl from the condition of BLOj at infinity: exponential decaying of ul

e

ΩO j

e

ˆ i (˜ divξˆ(ˆ cl j χj (ξˆ1j )Γ uej (ξˆL )), 0, ..., 0)T )dξˆ =

i.e. −

e

βj

u ˜ej (ξ L )dξˆ cl j =

βj

βj

ˆ Tj gj )1 dξ, (Γ

ˆ Tj gj )1 dξδl,2 , (Γ

(4.5.19)

where the upper 1 corresponds to the first component of the vector. index Note that βj u ˜ej (ξ L )dξ is negative due to the principle of maximum for e e e e c j and dˆ j = c j |ej |. Then we determine problem (4.5.10). Then we find c j = −ˆ l

e

l

l

l

j 0 the constants dl+1 from the condition of the exponential decaying of pBLO at l infinity. To this end we consider first the problem (4.5.11)-(4.5.13) without the last term in the equation (4.5.11), i.e.

0 0 ν∆ξ u ¯BLO − ∇ξ p¯BLO = f0 (ξ)δl,2 l l

+

e

e

e=ej , j=1,...,n

cl j {−ν∆ξ (χj (ξ1j )Γj (˜ uej (ξ L ), 0, ..., 0)T ) e

+(

e

+ ∇ξ (χj (ξ1j )ξ1j )} + e


p+r=l−1

0 ¯BLO = − divξ { divξ u l

∂ ej uej (ξ L ), 0, ..., 0)T }, (4.5.20) e χj (ξ1 ))Γj (˜ ∂ξ1j

e=ej , j=1,...,n

e

e

{cl j χj (ξ1j )Γj (˜ uej (ξ L ), 0, ..., 0)T }, ξ ∈ ΩO0 (4.5.21)

with the Dirichlet condition 0 u ¯BLO |∂ΩO0 = 0. l

e cl j

(4.5.22)

Here the constants are just defined by (4.5.18), (4.5.19) and satisfy the condition divξ { cel χi (ξ1e )Γj (˜ ue (ξ L ), 0, ..., 0)T )}dξ = 0, ΩO 0

e=ej , j=1,...,n

223

4.5. FLOWS IN TUBE STRUCTURES i.e.

e=ej , j=1,...,n

βj

e

u ˜ej (ξ L )dξcl j = 0. e

e

(4.5.23) e

Indeed, the choice of the constants cl j = −ˆ cl j and cˆl j from (4.5.19) and condition (4.5.4) give relation (4.5.23). 0 It is known (for example [120]) that there exists the unique solution {¯ uBLO , l BLO0 BLO0 stabilizes to zero at infinity (on every p¯l } of this problem such that u¯l BLO0 stabilizes on every branch of ΩO0 , associated with branch of ΩO0 ) and ¯l BLO0 ∞j own constant p ¯ ). These constants are defined uniquely up to ej , to its l 0 ∞1 one com mon additional constant, which we fix here by a condition ¯BLO l = 0. Then we define e

and

j 0 ∞j dl+1 = −¯ pBLO , l

(4.5.24)

0 0 uBLO =u ¯BLO , l l

(4.5.25)

e

e

j 0 0 pBLO = p¯BLO + dl+1 χj (ξ1j ) l l

(4.5.26)

on every branch of ΩO0 , associated with ej . 0 0 , pBLO } satisfies equations (4.5.11)-(4.5.13). Obviously, this pair {uBLO l l BLOj BLOj and pl , j = 1, ..., n, are not defined The boundary layer functions ul in the vicinity of O. Therefore we should change a little bit the formulas of ua and pa far from the nodes Oj , j = 0, ..., n. e Let ηj (x1j ) be a smooth function defined on each segment ej , let it be one ej e if |x1 − |ej |/2| ≥ |ej |/4 and let it be zero if |x1j − |ej |/2| ≤ |ej |/8. Let η(x) = ej µ ηj (x1 ) for each cylinder Bj and let η = 1 on each γjµ . Then we redefine ua and pa as

ua =

K l=2

pa =

K+1 l=2

µl {

uel (

e=ej , j=1,...,n

µl−2

n x − Oi xe,L i )η(x)}, uBLO ( )χµ (x) + l µ µ i=0

(4.5.27)

pel (x1e )χµ (x) +

e=ej , j=1,...,n

+

K+1 l=2

µl−1

n i=0

N i=1

i pBLO ( l

p2e (Oi )(1 − χµ (x))θi (x)

x − Oi )η(x)}, µ

(4.5.28)

The consequence of this redefinition is a small discrepancy in the right hand side of the equations of order O(exp(−c/µ)) with the positive constant c : now we have the relations

224


ν∆˜ ua − ∇˜ pa = f + Φ,

(4.5.29)

div u ã − (˜ ua , − ∇)˜ ua = Ψ, x ∈ B µ

(4.5.30)

with the Dirichlet condition u ã = g

(4.5.31)

on ∂B µ , where ΨH 1 (B µ ) = O(exp(−c/µ)), ΦL2 (B µ ) = O(µK−1+(s−1)/2 )

(4.5.32)

with a positive constant c and

Ψdx = 0,

(4.5.33)

Bµ

because ∂B µ (˜ ua , n)ds = ∂B µ (g, n)ds = 0. We are going to prove the estimate u − u ã H 1 (B µ ) = O(µK−1+(s−1)/2 ),

(4.5.34)

where c1 > 0 does not depend on µ. First we construct a function ϕ such that v = ∇ϕ is a solution of the equation div v = Ψ, x ∈ B µ ,

(v, n) = 0, x ∈ ∂B µ ,

i.e. let ϕ be a solution of the Neumann problem ∆ϕ = Ψ, x ∈ B µ ,

∂ϕ = 0, x ∈ ∂B µ . ∂n

The existence of this solution is provided by the condition B µ Ψdx = 0. As 2 2 µ the boundary belongs to C then ϕ ∈ H (B ). If we consider the solution with vanishing average B µ ϕdx = 0, then we apply the Poincare inequality for B µ (cf. Appendix 4.A.2): 1

µ

∀ϕ ∈ H (B ),

1 ϕ dx ≤ µ mesB µ B 2

(

Bµ

udx)

2

+

A

(∇ϕ)2 dx.

Bµ

where the constant A does not depend on µ. Thus we obtain the estimate ϕH 1 (B µ ) = O(exp(−c/µ)) and therefore we can use Agmon, Duglas and Nirenberg [2]theory and obtain the estimate ϕH 2 (B µ ) = O(exp(−c2 /µ)), where the positive constant c2 does not depend on µ. Therefore v ∈ H 1 (B µ ) and vH 1 (B µ ) = O(exp(−c2 /µ)).

225


The second step is the construction of such a function w = rot ψ that w = −v on ∂B µ . This construction is described in [87] and it has a ”local N nature.” We make a partition of unity 1 = k=1 δk on B µ , in such a way that all supports of δk have the diameters of order µ and satisfy to the condition of the existence of such a change of variables that the procedure of [87] ch.1, sect. 2 can be applied and the corresponding vector valued function ψk can be constructed for every support, i.e. rot ψk |∂B µ = δk v|∂B µ . All these supports have non-empty intersections ∆k with the boundary of B µ . Moreover we can make homothetic dilatations ξ = (x − Ak )/µ in 1/µ times with some centers Ak for every support in such a way that its image σk does not depend on µ. All these σk can be ”uniformed”, i.e. can be extended up to a finite number (independent of µ ) of the domains σ such that all other domains could be obtained from them by rotations and translations. The functions δk can be taken satisfying relation δk (x) = δ˜k ((x − Ak )/µ), where the functions δ˜k do not depend on µ. Thus the problem of construction of all vector valued functions ψk is reduced to a finite (independent of µ ) number of problems : construct such a vector valued function ψ˜α that rotξ ψ˜α |∂B µ = δ˜α v|∂B µ and take ψα (x) = µδ˜α ((x − Ak )/µ). In this case we can estimate rotξ ψ˜α H 1 (σ) ≤ Cα v(µξ + Aα )H 1 (σ) , with the constants Cα uniformly bounded by a constant C independent of µ. Therefore there exists a positive constant c1 independent of µ such that rot ψH 1 (B µ ) = O(exp(−c1 /µ)), and v + wH 1 (B µ ) = O(exp(−c1 /µ)). Therefore the relation holds true: Ua = u ã − (v + w) ∈ Hdiv=0 (B µ ) and − = −ν

=

Bµ

ν

s i=1

∂ϕ ∂U a , ) dx + ( ∂xi ∂xi

Bµ

(U a , (U a , ∇)ϕ)dx =

s ∂(˜ ua − (v + w)) ∂ϕ , ) dx + (˜ ua −(v+w), (˜ ua −(v+w), ∇)ϕ)dx = ( ∂xi ∂xi Bµ B µ i=1

(f, ϕ) dx +

Bµ

−

Bµ

Bµ

a

(Φ, ϕ) dx + ν

Bµ

((v + w), (˜ u − (v + w), ∇)ϕ)dx −

s

(

i=1

Bµ

∂(v + w) ∂ϕ , ) dx − ∂xi ∂xi

(˜ ua , (v + w, ∇)ϕ)dx,

226


0 µ for all ϕ ∈ Hdiv =0 (B ). Now we can apply the Poincare-Friedrichs ´ estimate (cf. section 4.A.2): 1 µ 2 ∀ϕ ∈ H0 (B ), ϕ dx ≤ A (∇ϕ)2 dx, Bµ

Bµ

where the constant A does not depend on µ. (Moreover, decomposing the domain to some subdomains with diameter of order µ, in such a way that each subdomain contains a part of the boundary ∂B µ , this estimate can be proved with the factor µ2 A instead of A). Applying this estimate as well as the a priori estimate for Navier-Stokes equation from [87],[188], we obtain: U a − uH 1 (B µ ) = O(µK−1 )

and

ua − uH 1 (B µ ) = O(µK−1 ),

(4.5.35)

where c3 > 0 does not depend on µ. This estimate can be improved in a standard way (see section 2.1) as ua − uH 1 (B µ ) = O(µK+(s−1)/2 ). Remark 4.5.1. If the right hand side function is defined on each cylinder Bjµ as f = fj (

x−O x − Oj e ) + χµ (x)Γj (fˆj (x1j ), 0, ..., 0)T , ) + f0 ( µ µ

(4.5.36)

where fˆj are sufficiently smooth functions, and for all γj , j = 0, ..., n, it is defined as earlier x − Oj ), µ then this case can be easily reduced to a previous type of right hand side function by a subtraction of a partial solution f = fj (

upartial = 0, ppartial = −

0

e

x1j

e Fj (x1j ) fˆj (t)dt = −F

on each cylinder Bjµ . Indeed if u and p is the solution of problem (4.5.1)-(4.5.3) e Fj (x1j ) is the solution with the right hand side (4.5.36) then the pair u, p−χµ (x)F of the problem with the right hand side x − Oj x−O e ) + χµ (x)Γj fˆj (xe1i ), 0, ..., 0)T − ∇(χµ (x)F Fj (x1j ), f˜ = fj ( ) + f0 ( µ µ

on each cylinder Bjµ . This right hand side function has a support concentrated in the vicinities of the nodes Oj , i.e. supp f˜ ⊂ supp (χµ − 1). Now we develop fˆj by the Taylor’s formula in vicinities of the nodes and obtain the set of problems with the right hand side of a form (4.5.5). It can be treated as above.

227


4.5.2

Tube structure with m bundles of tubes

Consider now the case of m different bundles of segments 1 m B1 = ∪nj=1 ej,1 , ..., Bm = ∪nj=1 ej,m .

We suppose that all common points of these bundles are end points of some segments of these bundles. Let the union of all bundles be connected. Consider the tube structure Bαµ associated with the bundle Bα . Now we do not require a base βˆjµ to be a part of ∂Bαµ in case if it corresponds to a common point of two bundles. Let µ B µ = ∪m α=1 Bα be a domain with C 2 −smooth boundary.

Figure 4.5.3. Tube structure with m (two) bundles of tubes. Consider the Navier - Stokes system of equations (4.5.1)-(4.5.3) for this B µ with the right hand side concentrated in the neighborhoods of the nodes (as above). Let us enumerate all nodes , i.e. all ends O1 , ..., ON of the segments and all segments e1 , ..., eM . For each ej introduce local co-ordinates xej , related to one of its ends. We seek the solution in a generalized form (4.5.27), (4.5.28), i.e.

u ã =

K l=2

µl {

e=ej , j=1,...,M

uel (

N x − Oi xe,L i )η(x)}, uBLO ( )χµ (x) + l µ µ i=1

(4.5.37)

228


K+1

pã =

µl−2

l=2

pel (xe1 )χµ (x) +

i=1

e=ej , j=1,...,M

+

K+1

µl−1

N

i pBLO ( l

i=1

l=2

N

p2 (Oi )(1 − χµ (x))θi (x)

x − Oi )η(x), µ

(4.5.38)

where the terms have the same form and sense as above: e

e

e

e

e

e

e

uej (ξ L ), 0, ..., 0)T , pl j (x1j ) = cl j x1j + dl j , ul j (ξ L ) = cl j Γj (˜ e

e

here cl j , dl j are scalar constants, u ˜ej is the solution of the problem (4.5.10) and the boundary value problems are stated for each domain ΩOi , related to Oi ; i.e. let Oi be one of the nodes ( one of the ends of ej1 , ..., ejqi ) and let the local coordinates for each of these segments are related to Oi ; cut all cylinders Bjµ1 , ..., Bjµq , associated with ej1 , ..., ejq at the distance of |ej1 |/2, ..., |ejqi |/2 respectively and consider the part of B µ which contains Oi ; then we extend this part substituting the deleted parts of the cylinders Bjµ1 , ..., Bjµq by the half-infinite cylinders havi ing the same cross-sections and orientations as Bjµ1 , ..., Bjµq (these half-cylinders i contain only the truncated parts of Bjµ1 , ..., Bjµq , but not the resting parts); then i after the homothetic dilatation of this constructed domain in 1/µ times with respect to Oi we obtain the domain ΩOi . Remark 4.5.2. All pe2 have a common value pe2 (Oi ) in Oi for all e with one of the ends Oi . The boundary layer problem is similar to (4.5.11)-(4.5.13): i i ν∆ξ uBLO − ∇ξ pBLO = fi (ξ)δl2 l l

+

e

e=ej , j=j1 ,...,jqi

e

{cl j {−ν∆ξ (χj (ξ1j )Γj (˜ uej (ξ L )), 0, ..., 0)T ) e

e

+ ∇ξ (χj (ξ1j )ξ1j )}+

+(

e


p+r=l−1

∂ ej uej (ξ L ), 0, ..., 0)T + e χj (ξ1 ))Γj (˜ ∂ξ1j

e

e

j dl+1 ∇ξ (χj (ξ1j ))},

i divξ uBLO = − divξ { 2


e

(4.5.39)

e

{cl j χj (ξ1j )Γj (˜ uej (ξ L ), 0, ..., 0)T }, ξ ∈ ΩOi (4.5.40)

229

4.5. FLOWS IN TUBE STRUCTURES with the Dirichlet condition i uBLO |∂ΩOi = 0 l

(4.5.41)

if Oi is not a ”boundary node,” i.e. if the distance from it to the support of the function g is of order of one, and with the boundary condition i uBLO |∂ΩOi = gi (ξ) l

(4.5.42)

if Oi is a ”boundary node,” i.e. if the distance from it to the support of the function g is of order of µ. e We obtain for cl j the equations of the type (4.5.4): e u ˜ej (ξ L )dξcl j = 0 (4.5.43) e=ej , j=j1 ,...,jqi

βj

if Oi is not a ”boundary node,” or


e

βj

u ˜ej (ξ L )dξcl j =

(gi , n) dξ.

(4.5.44)

∂ ΩOi ∩supp gi

if Oi is a ”boundary node,” n is the outside normal. e ej Now consider the problem of definition of cl j and dl+1 from (4.5.39)-(4.5.44). This problem is equivalent to the problem of definition of piecewise linear function Pl (x) defined on B (linear on every ej ) satisfying conditions (4.5.43),(4.5.44) e e where cl j = ∂P Pl /∂x1j , and satisfying conditions e

BLOi ∞j i dl j = −¯ pl−1 + dO l , BLOi ∞i i where p¯l−1 is a limit of the boundary layer function ¯BLO constructed as l−1 BLOi BLOi above in such a way that the pair u¯l−1 , p¯l−1 is a solution of the problem i (4.5.39)-(4.5.42) for l − 1 and without the last term of the equation (4.5.39); dO l is a constant independent of j . This condition can be rewritten in a form e

e

i ∞i dl j = −¯ pBLO + dl i1 . l−1

(4.5.45)

Let Φ be a function such that it is defined on each segment e, ∂Φ/∂xe1 = 0 in the vicinities of all nodes and such that in each node Oi its limit values Φej1 (Oi ), ..., Φejq (Oi ) for the segments ei1 , ..., eiq respectively satisfy to the rei lation (4.5.45), i.e. BLOi ∞j Φej (Oi ) = −¯ pl−1 + Φei1 (Oi ). ∂ 2 Pˆl ∂xe1 2

e

= 0 and that the constants cl j =

(4.5.46) ∂ Pˆl ∂xe1

satisfy relations ˆ (4.5.43),(4.5.44). Then we can reformulate the problem for Pl = Pl − Φ as We know that

230


−

e=ej , j=1,...,M

−

−

∂ Pˆl ∂ τˆ e dx = ∂xe1 ∂x1e 1

ρe

e

e=ej , j=1,...,M

ρe

O=Oi , boundary nodes

e

∂Φ ∂ τˆ e dx ∂x1e ∂x1e 1

(gi , n)dsτ (Oi ),

∂ Ω Oi

where the summation in the last term is developed over the boundary nodes, ρe = βe u ˜e (ξ L )dξ, τ is an arbitrary function of H 1 (B). Now the existence and uniqueness up to a constant of Pˆl is evident (by the Lax - Milgram lemma). The justification of series (4.5.37), (4.5.38) is the same as in the case of one bundle of segments B : a function U a ∈ Hdiv=0 (B µ ) is constructed in the same way as in section 1 and the following estimates are proved U a − uH 1 (B µ ) = O(µK ), ua − uH 1 (B µ ) = O(µK ).

4.6

(4.5.47)


The asymptotic analysis of the conductivity equation for finite rod structures first appeared as an auxiliary problem in the homogenization of the lattice (”skeletal”) structures [136], see also [16], Chapter 8, as well as [44]. The elasticity equations in an L-shaped finite rod structure was first considered in [94], and then in [145, 148] the complete asymptotic expansion of the displacement field in a large class of finite rod structures was constructed. The junction of two elastic rods with contrasting rigidities was studied in [168]. An arbitrary finite rod structure was studied in [123]. The two-dimensional cell problem for the conductivity equation and for the elasticity equations set on a rectangular lattice-like structure was studied in [78]. The complete asymptotic expansion was constructed and the homogenized equation coefficients were calculated explicitly: the effective conductivity (with the first corrector) and the effective shear modulus (which is small). The various junctions of plates, rods and three dimensional bodies were studied in [38],[80]-[82],[118],[178].

4.7

Appendices

4.A1. Appendix 1: estimates for traces in the pre-nodal domain Here we prove two estimates that were used in the proof of Theorem 4.1.1.

231

4.7. APPENDICES Lemma 4.A1.1. If the node x0 ∈ G0 then the estimate holds : ϕL2 (∂Π

++ x0 ∩∂Bµ )

√ ≤ C µ∇ϕL2 (Πx0 ) ,

where C does not depend on ϕ and µ.

Figure 4.A1.1. Pre-nodal domain Πx0 in Lemma 4.A1.1. Proof Make the change η = x/µ; one obtains for the images Ση of ∂Πx0 ∩ ∂Bµ++ and Πη of Πx0 the inequality ϕ2L2 (Ση ) ≤ C5 ϕ2H 1 (Πη ) , and from the classic Poincar´ é - Friedrichs inequality we obtain ϕ2L2 (Ση ) ≤ C6 ∇ϕ2L2 (Πη ) . Making the inverse change we get µ−s+1 ϕ2L2 (∂Π

++ x0 ∩∂Bµ )

≤ C6 µ−s+2 ∇ϕ2L2 (Π

x0 )

,

and it gives the desired estimate. Lemma 4.A1.2 The estimate holds : ϕ − < ϕ >Πx0 L2 (∂Π

++ x0 ∩∂Bµ )

where C does not depend on ϕ and µ.

√ ≤ C µ∇ϕL2 (Πx0 ) ,

232


Figure 4.A1.2. Pre-nodal domain in Lemma 4.A1.2. Proof In the integral i = |ϕ − < ϕ >Πx0 |2 dx ++ ∂ Πx0 ∩∂Bµ

make the change η = x/µ; we get i = µs−1 |ϕ − < ϕ >Πx0 |2 dη, Ση

where Ση is the image of the surface ∂Πx0 ∩ ∂Bµ++ and let Πη be the image of Πx0 under the transformation η = x/µ. Note that Ση is independent of µ. The estimate holds true in the dilated domain:

2

Ση

|ϕ − < ϕ >Πx0 | dη ≤ C

where C is independent of µ. Making the inverse change , we get

s ∂ϕ 2 ) dη, ( ∂η j Πη j=1

s s ∂ϕ 2 ∂ϕ 2 ) dx. ) dη = µ−s+2 ( ( Πx0 j=1 ∂xj Πη j=1 ∂ηj

Thus,

i ≤ Cµ

Πx0

s ∂ϕ 2 ) dx. ( ∂x j j=1

233

4.7. APPENDICES

By taking the square root of the two sides of the inequality, we obtain the assertion of the lemma. 4.A2 Appendix 2: the Poincar´ ´ e and the Friedrichs inequalities for a finite rod structure The goal of this appendix is to prove the Poincar´ é - Friedrichs inequality for a function u ∈ H1 (Bµ ), u = 0 on Σ0 . Below we consider domains which are connected and bounded open sets of IRs or IRs−1 with a piecewise - smooth boundary satisfying the cone condition. We suppose also that µ is sufficiently small. Lemma 4.A2.1. Let G1 , G2 be domains, G1 ⊂ G2 . Then for each function u ∈ H1 (G2 ) the estimate holds true ({u}1 − {u}2 )2 ≤ c (∇u)2 dx, G2

where c is a constant depending on G1 , G2 , and {u}i =

1 measGi

Gi

Figure 4.A2.1. Domains of Lemma 4.A2.1. Proof ({u}1 − {u}2 )2 =

≤

2 measG1

G1

1 measG1

G1

({u}1 − u)2 dx +

({u}1 − {u}2 )2 dx ≤

G1

({u}2 − u)2 dx

≤

u

dx.

234


≤

2 ( ({u}1 − u)2 dx + ({u}2 − u)2 dx) ≤ measG1 G1 G2 2c0 (∇u)2 dx, ≤ measG1 G2

c0 does not depend on u. Lemma is proved. Lemma 4.A2.2. Let Gµi = {x ∈ Rs | µx ∈ Gi }, where G1 , G2 are domains, G1 ⊂ G2 . Then for each u ∈ H1 (Gµ2 ) the estimate holds ({u}µ1 − {u}µ2 )2 ≤ cµ2−s (∇u)2 dx, Gµ 2

where c is independent of µ, but depends on G1 and G2 , {u}µi = Gµ udx/measGµi , i = 1, 2. i

Proof

({u}µ1

−

{u}µ2 )2

=

1 s µ measG1

Gµ 1

1 u dx − s µ measG2

Gµ 2

u dx

2

;

now the change of variables x′ = x/µ,

dx = µs dx′

yields ({u}µ1 − {u}µ2 )2 = ({u(µx′ )}1 − {u(µx′ )}2 )2 ≤ ≤c

(∇x u(µx′ ))2 dx′ =

G2

cµ2 µs

Gµ 2

(∇x u(x))2 dx

The lemma is proved. Lemma 4.A2.3. Let β be an (s−1) dimensional domain, β(a,b) be a cylinder (a, b) × β; denote µ β(a,b) = {x ∈ Rs | (

xs x2 , ..., ) ∈ β, µ µ

x1 ∈ (a, b)}.

Let d1 , d2 be constants independent of µ. Then ({u}µ(0,d1 µ)

−

{u}µ(0,d2 ) )2

d2 ≤ measβ

where u ∈ H1 (β(0,d2 ) ),

{u}µ(a,b)

=

1−s

µ

µ β(a,b)

µ β(0,d 2)

(

∂u 2 ) ∂x1

µ udx/measβ(a,b) .

dx,

235

4.7. APPENDICES

Figure 4.A2.2. Domains of Lemma 4.A2.3. Proof We obtain from Newton-Leibnitz formula y1 ∂u dx1 , u(y1 , x ˜1 ) − u(yo , x ˜1 ) = y0 ∂x1

x ˜1 = (x2 , ..., xs ).

Integrate this equality in x ˜1 ∈ β µ = {xǫRs−1 , µx ǫβ}, and in y0 from 0 to d1 µ. We obtain then:

| d1 µ

µ β(0,d

2)

udx − d2

µ β(0,d

1 µ)

udx |=|

d1 d2 µ||

0

d2

0

d1 µ

y1

y0

βµ

in y1 from 0 to d2

∂u d˜ x1 dx1 dy1 dy0 |≤ ∂x1

∂u || 1 µ ≤ ∂x1 L (β(0,d2 ) )

∂u || 2 µ . d1 d2 µ d2 µs−1 measβ|| ∂x1 L (β(0,d2 ) )

Divide this inequality by d1 d2 µs measβ; we obtain then √ ∂u d2 µ µ µ ||L2 (β(0,d || |{u}(0,d2 ) − {u}(0,d1 µ) | ≤ ). s−1 2) ∂x 1 µ measβ

Lemma is proved

236


Lemma 4.A2.4. Let β1 and β2 be two (s - 1) dimensional domains, β3 be s - dimensional domain. Denote βiµ = {x ∈ IRs−1 |

x ∈ βi } µ

β3µ = {x ∈ IRs |

(i = 1, 2),

x ∈ β3 }, µ

µ and let β˜i(a,b) be a cylinder obtained from the cylinder (a, b) × βiµ by some rotaµ µ tions and translations. Let β˜i(0,d ∩ β3µ = β˜i(0,h , where di , hi do not depend i) i µ) 7 7 2 µ 1 on µ. Let u ∈ H (B), where B = ( i=1 βi(0,di ) ) β3µ . Then 1 1 2 ( udx) , ( udx)2 ≤ µ µ µ µmeasβ ˜µ measβ˜1(0,d β β 3 3 ) (0,d ) 1

1

c1 ( udx)2 µ ˜µ measβ˜2(0,d β 2(0,d ) 2) 2

+

c2

(∇u)2 dx,

B

where c1 , c2 do not depend on µ.

Figure 4.A2.3. Domains of Lemma 4.A2.4. Proof. Denote 1 µ measβ˜1(0,d

1)

( βµ

1(0,d1 )

1 udx) as {u}d1,µ , then

µ 1µ 2 1µ 2 1 1 measβ˜1(0,d ({u}d1,µ )2 ≤ 2d1 µs−1 measβ1 (({u}h1,µ ) + ({u}d1,µ − {u}h1,µ ) ). 1)

237

4.7. APPENDICES According to Lemma 4.A2.3 the last expression is not more than h1 µ 2 s−1 2 2d1 µ measβ1 ({u}1,µ ) + 2d1 K (∇u)2 dx, µ β˜1(0,d

1)

Here constant K is independent of µ and d1 . The first term is estimated as follows: 2d1 µs−1

1µ 2 ({u}h1,µ ) ≤

measβ1

≤ 4d1 µs−1

measβ1 ({u}3,µ )2 + 4d1 µs−1 1 udx. here {u}3,µ = measβ µ β3µ 3 Then, applying Lemma 4.A2.2 we obtain

2d1 µs−1

1µ 2 ({u}h1,µ ) ≤ 4d1 µs−1

measβ1

+

4d1 µs−1

≤ 4d1

=

1 µ

measβ1 measβ3

+ Let {u}b2,µ

measβ1

1 µ measβ˜2(0,b)

µ β˜2(0,b)

measβ˜3µ ({u}3,µ )2

=

measβ1

cµ2−s

β3µ

measβ˜3µ

(4cd1 measβ1 )µ

1µ ({u}h1,µ − {u}3,µ )2 ,

measβ1

β3µ

({u}3,µ )2

+

(∇u)2 dx ≤

({u}3,µ )2

+

(∇u)2 dx.

udx. Then measβ3 µs ({u}3,µ )2

≤

h2 µ 2 2µ − {u}3,µ )2 ) + 2measβ3 µs ({u}h2,µ ≤ 2measβ3 µs ({u}2,µ

≤

2µ 2 2 2 )2 + 4measβ3 µs ({u}d2,µ − {u}h2,µ ) + ≤ 4mesβ3 µs ({u}d2,µ 4mesβ3 2 µmesβ˜2µ ({u}d2,µ )2 + + 2measβ3 cµ2 (∇u)2 dx ≤ µ d 2 mesβ2 β3

d2 +4measβ3 µ µ1−s K measβ2 s

2

µ β˜2(0,d

(∇u) dx

1 4measβ3 ≤ µ( µ d2 measβ2 measβ˜2(0,d

4measβ3 +( d2 K measβ2

+

2µcmeasβ3

µ β˜2(0,d

2)

2)

+

2)

(

2cmeasβ3 )

udx)2 +

B

(∇u)2 dx).

β3µ

(∇u)2 dx ≤

238


Lemma is proved with the following values of constants : c1 = max{16

c2 = max{4d1 measβ1 (

4measβ3 }, d2 measβ2

d1 measβ1 , d2 measβ2

4d2 K +3c)+2Kd21 , 4measβ3 d2 K/measβ2 +2cmeasβ3 }, measβ2

(µ < 1). Lemma 4.A2.5. Let u ∈ H 1 (Gµ ), Gµ = {x ∈ IRs | µx ∈ G} and G is a domain in IRs . Then the estimate takes place 1 ( udx)2 + µ2 c||∇u||2L2 (Gµ ) . ||u||2L2 (Gµ ) ≤ measGµ Gµ

Here constant c depends on G. Proof. Change the variables ξ = µx , dx = µs dξ. The classical Poincare inequality yields 1 2 s 2 s ( u(µξ)dξ)2 + ||u||L2 (Gµ ) = µ u (µξ)dξ ≤ µ measG G G

+µs c

(∇ξ u(µξ))2 dξ

=

G

1 ( u(x)dx)2 measGµ Gµ

+ µ2 c

(∇x u(x))2 dx.

Gµ

Lemma is proved. µ Lemma 4.A2.6. Let u ∈ H 1 (β(0,d) ) then ||u||2L2 (β µ

(0,d)

)

≤

4 ( u(x)dx)2 + 8d2 (∇u)2 dx µ µ µ measβ(0,d) β(0,d) β(0,d)

for all sufficiently small µ. Proof For any y1 , y0 ∈ R, x ˜1 ∈ Rs−1 we obtain (u(y1 , x ˜1 ))2 ≤ 2(u(y1 , x ˜1 ) − u(y0 , x ˜1 ))2 + 2u2 (y0 , x ˜1 ) ≤ ≤ 2(u(y0 , x ˜1 )) + 2( 2

y1

∂u(x) dx1 )2 ≤ ∂x1

yo

2

≤ 2(u(y0 , x ˜1 )) + 2d

Integrate the inequality in x˜1 over µ.

0

µ β(0,d) ,

d

(

∂u 2 ) dx1 ∂x1

in y1 from 0 to d, in y0 from 0 to

239

4.7. APPENDICES

µ

µ β(0,d)

u2 dx ≤ 2d

µ β(0,µ )

u2 dx + 2d2 µ

µ β(0,d)

(

∂u 2 ) dx. ∂x1

µ with constant cP µ2 gives: The Poincare´ inequality for β(0,µ)

µ

µ β(0,d)

2d 2 2 ( udx) + 2dc µ (∇u)2 dx+ P µ µ µs measβ β(0,µ) β(0,d)

u2 dx ≤

+2d2 µ

µ β(0,d)

(

∂u 2 ) dx. ∂x1

So, µ

+2µd(d + µcP )

µ β(0,d)

µ β(0,d)

u2 dx ≤ 2d(µs measβ({u}µ(0,µ) )2 +

(∇u)2 dx) ≤ 4dµs measβ(({u}µ(0,d) )2 + ({u}µ(0,µ) −

{u}µ(0,d) )2 ) + 4d2 µ

µ β(0,d)

(∇u)2 dx

for all µ ≤ d/cP . Lemma 4.A2.3 yields

µ

µ β(0,d)

u2 dx ≤

4µ 2 2 ( udx) + 8d µ (∇u)2 dx. µ µ µ measβ(0,d) β(0,µ) β(0,d)

Here {u}µ(a,b)

=

1 µ measβ(a,b)

µ β(a,b)

udx.

Lemma 4.A2.6 is proved. Lemma 4.A2.6 admits the following generalization: µ µ Lemma 4.A2.7. Let the domain β˜(a,b) be obtained from β(a,b) by means of rotation and translation, then the estimate holds true 4 2 u2L2 (β˜µ ) ≤ ( udx) + c (∇u)2 dx, 3 (0,d) measβ˜µ β˜µ β˜µ (0,d)

(0,d)

(0,d)

where c3 is independent of d and µ, µ is sufficiently small. Theorem 4.A2.1. (Poincar´ ´ e inequality for the rod structure) Let β1 , ..., βr be (s − 1)−dimensional domains, β3,1 , ..., β3,p be s−dimensional doµ mains, βiµ = {x ∈ IR(s−1) |x/µ ∈ βi }, i = 1, ..., r, β3,j = {x ∈ IR(s−1) |x/µ ∈

240


µ β˜i,(a,b) be cylinders, obtained from the cylinders (a, b)×βiµ µ µ by means of rotations and translations and let β˜3,j are obtained from β3,j by translations only. Let di , hi do not depend on µ. We assume that the intersecµ µ tion of closures of the sets β˜i,(0,d and β˜j, (0,dj ) is empty and that the intersection i) µ µ ˜ ˜ of closures of the sets β3,i and β3,j is also empty if i = j. Moreover we assume that for each pair i, j either the intersection of closures of the sets β˜µ and

β3,j },

j = 1, ..., p,

i,(0,di )

µ µ µ µ is empty or β˜i,(0,d ∩ β˜3,j = β˜i,(0,h . β˜3,j i) i µ) µ µ r ˜ ˜ Let the set B = ∪i=1 βi,(0,di ) ∪ ∪pj=1 β˜3,j be connected. Then the estimate holds true 1 ( udx)2 + A (∇u)2 dx, u2 dx ≤ ˜ ˜ ˜ ˜ measB B B B

where A does not depend on µ, µ is sufficiently small. µ If there exists such a cylinder β˜r,(0,d that β˜µ u dx = 0, then r)

˜ B

Proof Suppose that β˜µ

u2 dx

≤

A

˜ B

r,(0,dr )

(∇u)2 dx.

u dx = 0. We shall prove that

r,(0,dr )

˜ B

u2 dx

≤

A

˜ B

(∇u)2 dx.

First we deduce from lemmas 4.A2.6, 4.A2.7 that for all i = 1, ..., r 4 2 2 ( udx) + A1 (∇u)2 dx, u dx ≤ ˜µ µ ˜µ ˜µ meas β β β β˜i,(0,d i,(0,di ) i,(0,d ) i,(0,d ) )

i

i

i

(when i = r the first term vanishes) and that for all j = 1, ..., p 1 2 2 2 ( udx) + µ c (∇u)2 dx, u dx ≤ measβ˜µ β˜µ β˜µ β˜µ 3,j

3,j

3,j

3,j

where A1 , c are independent of µ. Fix any i (respectively j ). Define the initial µ µ set as the set β˜i,(0,d (respectively the set β˜3,j ). i) ˜ The set B is connected. Therefore there exists the chain c of sets µ µ µ β˜3,j , β˜iµ1 ,(0,di ) , β˜3,j , β˜iµ2 ,(0,di ) , ..., β˜3,j , 1 2 q 1

2

such that each two consecutive sets (units) have the nonempty intersection, µ such that the intersection of the first of these units with the initial set β˜i,(0,d i) is also not empty and µ such that the intersection of the last of these units β˜3,j with the set of the q µ is also not empty. vanishing average β˜ r,(0,dr )

µ (In the case of the initial set β˜3,j the chain starts from β˜iµ1 ,(0,di

1)

directly.)

241

4.7. APPENDICES

Figure 4.A2.4. A chain of the proof of Theorem 4.A2.1. µ µ Indeed, at the first step we can add to the initial set all units β˜3,l and β˜l,(0,d l) which have the nonempty intersection with the initial set; then at the second step we add all units which have the nonempty intersection with the units obtained at the first step and so on. After the finite number of steps we shall either obtain ˜ or come to a contradiction to the connectivity of B ˜ . We can the whole set B also suppose that all units are different in the chain C because if there is a unit repeated twice then we can exclude all the units between these two repetitions without loss of connectivity of the chain. Thus there are not more than r + p units in the chain. Applying lemma 4.A2.4 q + 1 times maximum we obtain the inequality (max c1 )q+1 1 ( udx)2 + ( udx)2 ≤ µ ˜µ measβ˜r,(0,d measβ˜ β˜ β r,(0,dr ) r) q max c2 (max c1 )q (∇u)2 dx, ˜ B

i=0

where β˜ is the initial set of the chain, max c1 and max c2 are the maximal values µ µ , β˜2,(0,d ). of the constants c1 > 1 and c2 of lemma 4.A2.4 for all pairs (β˜1,(0,d 1) 2) udx = 0 we obtain: Since β˜µ r,(0,dr )

1 ( udx)2 measβ˜ β˜

Thus

≤

β˜

u2 dx

max c2

≤

(max c1 )q+1 − 1 max c1 − 1

A2

˜ B

(∇u)2 dx,

˜ B

(∇u)2 dx.

242


where A2 is some constant independent of µ, and, therefore,

˜ B

2

u dx

≤

r

2

µ β˜i,(0,d

i=1

Let now Dr =

˜ B

+

r)

2

µ β˜3,j

j=1

i)

µ β˜r,(0,d

u dx

≤

(p+r)A2

˜ B

(∇u)2 dx.

udx = 0, then

2

˜ B

u dx

p

1 ( udx)2 ≤ ˜ B˜ measB ≤ (p + r)A2 (∇u)2 dx.

−

u dx

(u − Dr )2 dx

˜ B

− z) dx. Let F (z) = B˜ (u − z)2 dx, then ˜ z − 2 F ′ (z) = 2 measB udx

Indeed calculate minz∈R

˜ (u B

2

˜ B

and minz∈R F (z) =

˜ B

u2 dx

−

1 ( udx)2 . ˜ B˜ measB

Theorem is proved. Let now x0 ∈ G0 , ( G0 be defined in section 4.1). Consider the parallelepiped Πx0 = (−1, −d0 µ) × (x02 − d0 µ, x02 + d0 µ) × ... × (x0s − d0 µ, x0s + d0 µ). Assume that ϕ ∈ H1,0 (Bµ ). Then we can extend ϕ as a zero onto Πx0 \Bµ . So this extension ϕ belongs to H 1 (Bµ ∪ Πx0 ). Due to Theorem 4.A2.1 the estimate holds ϕ2 dx ≤ A (∇ϕ)2 dx, Bµ ∪Πx0

Bµ ∪Πx0

and hence

Bµ

ϕ2 dx ≤ A

(∇ϕ)2 dx.

Bµ

Thus the Poincar´ é - Friedrichs inequality in Bµ with the constant independent of µ is proved. Formulate it as a theorem. Theorem 4.A2.2. Let ϕ ∈ H1,0 (Bµ ). Then ϕ2 dx ≤ A (∇ϕ)2 dx. Bµ

Bµ

where the constant A does not depend on µ, µ is sufficiently small. All results of this Appendix were proved in [144]. 4.A3 Appendix 3: the Korn inequality for the finite rod structures. Here we prove the Korn inequality for the finite rod structures such that every pair of the intersecting constitutive cylinders is star-wise with respect to

243

4.7. APPENDICES

some ball of the radius of order µ. The proof in a more general case will be given in the appendix to the next chapter. Lemma 4.A3.1 [129]. Let Ω be a bounded domain star-like with respect to the ball QR1 = {x : |x| < R1 } let the diameter of Ω be R, u = (u1 , . . . , us ) ∈ (H 1 (Ω))s . Then ∇u2L2 (Ω) ≤ C1 ( RR1 )s+1 e(u)2L2 (Ω) + C2 ( RR1 )s ∇u2L2 (QR ) , where C1 , C2 de1 pend on s only, 2 s ∂ui ∂uj 1 2 dx, + e(u)L2 (Ω) = ∂xi 4 Ω i,j=1 ∂xj

∂uj 1 ∂ui . + e(u) is the matrix s × s with elements ∂xi 2 ∂xj be a subspace Theorem 4.A3.1. Let Ω be a domain of Lemma 4.A3.1, let H 1 s of (H (Ω)) ; let QR1 belongs to a subdomain Ω1 ⊂ Ω such that there exists a the following estimate constant C3 depending on s only such that for any u ∈ H, holds : ∇u2L2 (Ω1 ) ≤ C3 e(u)2L2 (Ω1 ) .

Then the inequality holds for all u ∈ H s+1 R e(u)2L2 (Ω) ∇u2L2 (Ω) ≤ C4 R1

(4.A3.1)

where C4 depends on s only. Proof Applying Lemma 4.A3.1 and estimating ∇u2L2 (QR

1

)

≤ ∇u2L2 (Ω1 ) ≤ C3 e(u)2L2 (Ω1 ) ≤ C3 e(u)2L2 (Ω) R1 R

≤ 1. The theorem is proved. µ µ . Consider now a thin rectangle Gµ that is (0, 1) × − , 2 2 Corollary 4.A3.1 (for Theorem 4.A3.1) There exists C6 > 0 such that for any µ > 0 small enough, for any u ∈ (H 1 (Gµ ))s , such that u|x1 =0 = 0, the estimate holds true : ∇u2L2 (Gµ ) ≤ C6 µ−3 e(u)2L2 (Gµ ) . Proof µ µ . Let us consider the square Ω1 = (0, µ) × − , 2 2 1 For any u ∈ H (Gµ ) such that u|x1 =0 = 0, ∇u2L2 (Ω1 ) ≤ C7 e(u)2L2 (Ω1 ) . Applying now theorem Theorem 4.A3.1 with the disk 8 5 µ2 µ 2 + x22 < QR1 = x1 , x2 ∈ R2 ; x1 − 4 2

we get (4.A3.1 ) with C4 = C1 + C2 C3 because

we obtain the assertion of the corollary.

244

CHAPTER 4. FINITE ROD STRUCTURES Theorem 4.A3.2 Let Ω be a domain of Lemma 4.A3.1 and let u be a vector-

valued function of H 1 (Ω) such that for any rigid displacement η, ∂ui ∂ηi dx = 0. . Ω ∂xj ∂xj

Ω

(∇u, ∇η)dx =

i,j=1

Then (4.A3.1 ) holds with the constant C4 depending on s only. Proof (∇(u − η), ∇η)dx = 0, for Consider a rigid displacement η, such that QR1

all rigid displacements η.

Then ∇(u−η)2L2 (Ω) = ∇u2L2 (Ω) −2

Ω

(∇u, ∇η)dx+∇η2L2 (Ω) = ∇u2L2 (Ω) +

∇η2L2 (Ω) ≥ ≥ ∇u2L2 (Ω) . Applying Lemma 4.A3.1 to u − η, we get s s+1 R R ∇(u − η)2L2 (QR ) e(u − η)2L2 (Ω) + C2 ∇(u − η)2L2 (Ω) ≤ C1 1 R1 R1

Applying now the Korn inequality for a function orthogonal to all rigid displacements in a ball (for QR1 ) [129], we have : ∇(u − η)2L2 (QR

1

)

≤ C5 e(u − η)2L2 (QR

1

)

with C5 depending only on s. So, finally, ∇u2L2 (Ω) ≤ ∇(u − η)2L2 (Ω) ≤ s s+1 R R C5 e(u − η)2L2 (QR ) ≤ e(u − η)2L2 (Ω) + C2 ≤ C1 1 R1 R1 s+1 s+1 R R e(u)2L2 (Ω) e(u − η)2L2 (Ω) = (C1 + C2 C5 ) ≤ (C1 + C2 C5 ) R1 R1

because e(u − η) = e(u). The theorem is proved. Consider the set A of finite rod structures Bµ such that (1) (2) - for any pair of cylinders Bµ and Bµ constituting Bµ and having common points in one of their bases, there exists a ball Qc0 µ of the radius c0 µ (c0 is (1)

(2)

independent of µ) such that it belongs to the union B µ ∪ B µ and such that this union is star-wise with respect to the ball Qc0 µ ; and - for any cylinder with a base belonging to ∂1 Bµ there exists a ball Qc0 µ of the radius c0 µ in a µ-vicinity of ∂1 Bµ such that this cylinder is a star-wise domain with respect to this ball Qc0 µ (if such cylinder is associated with an additional cube of Remark 4.1.1, then we require the cylinder with the cube to be star-wise domain with respect to a ball Qc0 µ belonging to the cube).

245

4.7. APPENDICES

It is evident that in the case of dimension 2 all the finite rod structures belong to A, and in the case of dimension 3 if all the bases of cylinders are star-wise with respect to some ball of the radius c0 µ then Bµ belongs to A. More general finite rod structures will be studied in Appendix 5.A2.

Figure 4.A3.1. To the proof of the Korn inequality Theorem 4.A3.3.Let Bµ ∈ A. There exists C7 > 0 and r ∈ R such that for any positive µ small enough, for any u ∈ (H 1 (Bµ ))s , u|∂1 Bµ = 0 (or u| 0 = 0), the estimate holds true : ∇u2L2 (Bµ ) ≤ C7 µ−r e(u)2L2 (Bµ ) .

246


Proof. Apply lemma 4.A3.1 to all cylinders and their balls Qc0 µ of the definition of (0) set A. Estimate ∇u2 2 (0) for a ball Qc0 µ in the µ-vicinity of ∂1 Bµ (i.e. 0 ) by C8 e(u)2 2 L

L (Qc0 µ )

(0) (Qc0 µ )

with a constant C8 > 0 independent of µ as in Theorem

(1)

4.A3.1. Let now Qc0 µ be another ball of the definition of A. Consider a chain of (1) (N ) (1) (0) (1) cylinders Bµ , . . . , Bµ relating Qc0 µ and Qc0 µ in such a way that Bµ contains (1) (N ) (0) (i) (i+1) Qc0 µ and Bµ contains Qc0 µ and every two consecutive cylinders Bµ , Bµ (i+1)

contain a ball Qc0 µ

(i+1)

such that Qc0 µ

(i+1)

(i)

(i)

(i+1)

star-wise domain with respect to Qc0 µ . We have by Lemma 4.A3.1,

∇u2L2 (Q(1) ) ≤ ∇u2L2 (B (1) ) ≤ c0 µ

µ

≤ C9 µ−(s+1) (e(u)2L2 (B (1) ) + ∇u2L2 (Q(2) ) ) ≤ µ

−(s+1)

≤ C9 µ

c0 µ

(e(u)2L2 (B (1) ) + µ

+C C9 µ−(s+1) (e(u)2L2 (B (2) ) + ∇u2L2 (Q(3) ) )) ≤ . . . µ

−N (s+1)

≤ C10 µ

(e(u)2L2 (Bµ )

c0 µ

+

(i+1)

and B µ ∪ B µ

⊂ B µ ∪ Bµ

∇u2L2 (Q(0) ) ) c0 µ

≤

≤ C11 µ−N (s+1) e(u)2L2 (Bµ ) , where C8 , C9 , C10 , C11 > 0 do not depend on µ. Theorem is proved.

is a

Chapter 5

Lattice Structures Here we consider the so called lattice structures, i.e. finite rod structures of Chapter 4 with a great number of elementary rods. Its definition is given at section 5.1. As in the previous chapter we first study these structures by L-convergence technique for an elliptic equation in section 5.2 and for dynamic problems in section 5.3. We start the study from the simplest (cross-like) rectangular lattice and then (in section 5.3) consider the general case. In section 5.4 the applicability of the L-convergence technique to the elasticity problem is discussed. Then we develop a refined analysis of fields, including asymptotic expansion of elasticity equation based on the expansions of Chapter 2 and Chapter 4. We apply first the analysis of Chapter 4 to obtain some limit differential equations at the axial segments of the elementary rods. Then we apply the homogenization method to find the asymptotic expansion as the period ε tends to zero. Finally, we consider the L-convergence of random lattices. We generalize the L-convergence tools in this case and formulate a result of the convergence for such structures with high probability (section 5.7).

5.1

Definition of lattice structure

The lattice-like domains simulate some widely adopted engineering constructions such as frameworks of houses, trusses of bridges, industrial installations, supports of electric power lines, spaceship grids, etc. as well as some capillary or fissured systems. The methods of solving the system of equations of the strength of materials theory (structural mechanics methods) presently used for calculating the frameworks have the drawback that the number of equations increases with number of nodes of lattice. This greatly impedes the calculation of lattices with large number of nodes, especially when investigating non-stationary and nonlinear processes. The homogenization techniques and splitting principle for the homogenized operator proposed in [134]- [136], [16] allowed essentially simplify the process of calculation of such structures.

247

248

CHAPTER 5. LATTICE STRUCTURES

Definition 5.1.1 Let β1 , ..., βJ be bounded domains in IRs−1 , (s = 2, 3) with a piecewise smooth boundary satisfying the strong cone condition, let Bj (j = 1, ..., J) be cylinders, suppose Bj = {x ∈ IRs | (x2 /µ, ..., xs /µ ∈ βj , x1 ∈ IR}, α ˜ with and Bh,j is the cylinder obtained from Bj by orthogonal transformation Π the matrix αT , α = (αil ) and by translation by h = (h1 , ..., hs )T . Let eα h be an s-dimensional vector obtained from the vector i(h, α) of the axis Ox1 with ˜ of IRs with matrix the beginning at the point O by orthogonal transformation Π T α and by translation by h = (h1 , ..., hs ). Let B be a union of all vectors eα h 2 when α belongs to a set ∆ ⊂ IRs and h to a set Hα ⊂ IRs , and these sets are independent of ε, µ. Let B be such that any two segments eα h can have only one common point which is the end point for both segments. The end points of eα h are called nodes. We assume that the system of the segments B is connected and periodic in all the xi with period 1, with periodicity cube Q = {x ∈ IRs | − 1/2 ≤ xi ≤ 1/2} αI 1 intersecting a finite number of segments from B : eα h1 , ..., ehI . To each of the αi αi ˜ αi segments ehi , (i = 1, ...I), we assign the cylinder Bhi ,ji and denote by B hi ,ji αi the part of Bhi ,ji contained between the two planes passing through the ends of ˜ αi segment eα hε and perpendicular to it (we assume that the bases belong to Bhi ,ji ). Denote Bξ,µ , the set of interior points of the periodic continuation (with the ˜ αi , i = 1, ..., I, periodicity cube (0, 1)s ) to IRs of the union of the cylinders B hi ,ji while assuming that the intersection of the set Bξ,µ and the cube Q is connected. By lattice structure Bε,µ we shall mean the set {x ∈ IRs | x/ε ∈ Bξ,µ } produced from Bξ,µ by homothetic contraction of space in 1/ε times. We will assume that the lattice structure has a piecewise smooth boundary, satisfying the cone condition. s ˜α ˜α The maximal subsets of the cylinders B h,j,ε = {x ∈ IR | x/ε ∈ Bh,j } for ˜α which any cross-section by the plane perpendicular to the generatrix of B h,j,ε is free of points of other cylinders we call sections. By S0 we denote the set of sections and let c0 εµ > d, where d is the maximal diameter of the connected subsets Bε,µ \S0 , with c0 being independent of ε and µ. By eα hε we will mean the α segment {x ∈ IRs | x/ε ∈ eα h } produced from eh by homothetic contraction of space in 1/ε times. ˜ α,+ (B ˜ α,++ ) be a part of B ˜ α , contained between the planes spaced Let B h,j,ε h,j,ε h,j,ε + ˜ α . Let Bεµ by c0 εµ (by (c0 + 1)εµ) from the bases of B be a union of all h,j,ε α,+ α,++ ++ ˜ ˜ ¯ ++ B and let B be a union of all B . The domain B εµ \Bεµ is called εµ h,j,ε h,j,ε the nodal domain.

5.2.

L-CONVERGENCE HOMOGENIZATION OF LATTICES

249

Figure 5.1.1. Domain G independent of small parameters and its intersection with the rectangular lattice.

5.2

L-convergence homogenization of lattices

5.2.1

L-convergence for the simplest lattice

We start from the example of the simplest model of lattice-structures: twodimensional rectangular lattice. We prove all necessary auxiliary results for this simplest lattice (even if these results are some particular cases of the results of previous chapters); so the first section can be read absolutely independently of others. We announce here below some generalizations, proved later. Definition 5.2.1 The union 2 Bε,µ = ∪+∞ k=−∞ ( {(x1 , x2 ) ∈ IR | | x2 − kε | < εµ/2 }

∪ {(x1 , x2 ) ∈ IR2 | | x1 − kε | < εµ/2 } ) is called the two-dimensional rectangular lattice. Thus the rectangular lattice is a union of thin strips of the width εµ stretched in each coordinate direction and forming the ε− periodic system in each dimension. We also denote 2 j = ∪+∞ Bε,µ k=−∞ {(x1 , x2 ) ∈ IR | | x3−j − kε | < εµ/2 }

the unions of horizontal (j = 1) and vertical (j = 2) strips, so

250


1 2 Bε,µ = Bε,µ ∪ Bε,µ .

Let G be a domain with the boundary ∂G ∈ C ∞ which is independent of ε and µ. In the domain Bε,µ ∩ G we consider the equation −div ( A grad uε,µ ) = f (x) ,

(5.2.1)

with the boundary conditions ( A grad uε,µ , n ) = 0, f or x ∈ ∂Bε,µ ∩ G,

(5.2.2)

¯ε,µ ∩ ∂G, uε,µ = 0, f or x ∈ B

(5.2.3)

here x = (x1 , x2 ), A = (aij ) is a constant (2 × 2)−matrix independent of ε and µ , A = AT > 0 , i.e. aij = aji and A is positive, f ∈ C 1 (G). Problems (5.2.1)-(5.2.3) simulate a steady state heat field in the lattice structure Bε,µ ∩ G under the conditions of thermal insulation on the boundary of G with A being the heat conductivity tensor for the material of which the framework is made (”cut out”). For aij = δij D (5.2.1)-(5.2.3) may be interpreted also as a problem of diffusion of a substance in a fissured rock filled with water or oil, with D being the diffusion coefficient of the substance in the fluid. In this case, the framework Bε,µ models a rectangular system of cracks εµ wide, filled with water. This model is very approximate since the system of cracks in a rock has a less regular structure . The model of a random framework (section 5.9) is more suitable for its description. Numerical solution of problems (5.2.1)-(5.2.3), with ε 0, S ≥ 0) are constants A = (aij ) is a constant (2×2)−matrix independent of ε and µ , A = AT > 0 ,i.e. aij = aji and A is positive, ¯ × R+ ), f (x, 0) = ∂f |t=0 = 0. f ∈ C ∞ (G ∂t As above ∂G is smooth enough: ∂G ∈ C 3 . Definition 5.3.1 Let uε,µ (x, t) be a sequence of functions from L2 ((Bε,µ ∩ G) × [0, T ]) , and u0 (x) ∈ L2 (G × [0, T ]). One says that uε,µ Lt −converges to u0 (x, t) at (Bε,µ ∩ G) × [0, T ] if and only if uε,µ |t=0 = 0,

uε,µ − u0 L2 ((Bε,µ ∩G)×[0,T ]) → 0, mes(Bε,µ ∩ G)

(ε, µ → 0).

(5.3.5)

5.3.

259

NON-STATIONARY PROBLEMS

For the solution of a homogenized problem R

∂u0 ∂ 2 u0 − div ( Aˆ grad u0 ) = f (x, t) , x ∈ G, t > 0, + S ∂t ∂t2

u0 |∂G = 0,

(5.3.6) ∂u0 (5.3.7) |t=0 = 0, u0 |t=0 = 0, ∂t one obtains the following theorem of Lt − convergence,proved in [16]: Theorem 5.3.1 The solution uε,µ of the initial problem Lt −converges to the solution u0 of the homogenized problem and

√ uε,µ − u0 L2 ((Bε,µ ∩G)×[0,T ]) √ = O( ε + µ), mes(Bε,µ ∩ G)

(ε, µ → 0).

(5.3.8)

Denote

uF Lt((Bε,µ ∩G)×[0,T ]) = * supt0 ∈[0,T ] supϕ|t=t0 ∈H01 (Bε,µ ∩G) |

((R

Bε,µ ∩G

(A∇u , ∇ϕ) dx|/(ϕH 1 (Bε,µ ∩G)

∂u ∂2u + S ) , ϕ) + ∂t ∂t2

6

mes(Bε,µ ∩ G)) .

Introduce FLt-convergence. Definition 5.3.2 Let uε,µ (x, t) be a sequence of functions from H01 (Bε,µ ∩G) for almost all t ∈ [0, T ], such that

∂uε,µ ∂ 2 uε,µ |t=t0 ∈ L2 (Bε,µ ∩ G) |t=t0 , ∂t ∂t2

for almost all t0 ∈ [0, T ]. One says that uε,µ FLt - converges to zero if and only if uε,µ F Lt((Bε,µ ∩G)×[0,T ]) → 0,

(ε, µ → 0).

(1)

The theorems on FLt-convergence of the difference uε,µ − uε,µ to zero could be formulated and proved as in [16] (see the next subsection).

5.3.2

Lattices: general case

Consider now the general lattices of Definition 5.1.1. Let G be a bounded domain in IRs with a piece-wise smooth boundary ∂G, (satisfying the cone condition). Consider the problem s

R(x, t)

∂u ∂2u − ∂/∂xi ( Ai (∇x uε,µ , x, t)) = + S(x, t) 2 ∂t ∂t i=1

260

CHAPTER 5. LATTICE STRUCTURES f (x, t) , f or x ∈ Bε,µ ∩ G, t > 0, ∂uε,µ /∂ν = ni Ai (∇x uε,µ , x, t) = 0, f or x ∈ ∂Bε,µ ∩ G,

(5.3.9)

(5.3.10)

¯ε,µ ∩ ∂G, uε,µ = 0, f or x ∈ B

(5.3.11)

∂uε,µ |t=0 = 0, ∂t

(5.3.12)

uε,µ |t=0 = 0,

where R(x, t) and S(x, t) are n × n matrix-valued functions, u(x, t), f (x, t), Ai (y, x, t) (i = 1, ..., s) are n −dimensional vector-valued functions, ∇x u =(∂uk / ∂xl ) is an n × s matrix, y =(ykl )is an n × s matrix-argument,f, R,S are assumed to be continuously differentiable, and Ai to be twice continuously differentiable, and ni (x) is the cosine of the angle between the axis Oxi and the exterior normal vector n(x). We assume that R > 0. We can consider a ”parabolic” case when R = 0 and S > 0; if so, the second initial condition (5.3.12) should be cancelled. By formally taking R = S = 0, we can consider the steady state problem. Then both conditions (5.3.12) are cancelled. We suppose that problem (5.3.9)-(5.3.12) has the solution, such that for each n−dimensional vector-valued function ϕ(x, t) which belongs to the space H 1 (Bε,µ ∩ G) for each fixed value of t and equal to zero on the boundary ∂G, the integral

((R

Bε,µ ∩G

∂uεµ ∂ 2 uεµ ) , ϕ) + (A∇uεµ , ∇ϕ) dx + S ∂t ∂t2

is defined and is equal to

(f (x, t) , ϕ) dx.

Bε,µ ∩G

We shall describe the procedure of constructing the homogenized operator. It is related to resolving of the cell problem by L-convergence method of section 4.1.2. To each segment e ⊂ B , we assign a collection of n−dimensional vectors (assuming their existence) X0 (y, x, t), ..., Xs (y, x, t) ∈ C 2 , which satisfies the following four conditions: 1) Ai (X1 (y, x, t), ..., Xs (y, x, t), x, t) νij = 0, where j = 1, ..., s − 1, and the vectors ν 1 = (ν11 , ..., νs1 ) , ... , ν s−1 = (ν1s−1 , ..., νss−1 )

5.3.

261


form the basis in (s − 1)−dimensional space orthogonal to e, y = (y1 , ..., ys ) is n × s−matrix-parameter. 2) Let the segments e1 , ..., eq have a common end point; then q s

e Ai (X1 j (y, x, t), ..., Xsej (y, x, t), x, t) ηij mesβ˜j = 0,

j=1 i=1

where η1j , ..., ηsj are the direction cosines of ej , and β˜j is a cross-section of a cylinder with the axis ej by the hyperplane orthogonal to this axis. 3) The vector-valued function defined on each segment e ⊂ B by s i=1

(Xie (y, x, t) − yi ) ξi + X0e (y, x, t)

(for ξ ∈ e ) is a continuous 1-periodic function of (ξ1 , ..., ξs ) ∈ B . 4) The functions equal to Xie (y, x, t) for ξ ∈ e ⊂ B , which are piecewise constant in the variables ξ, are 1-periodic. ˆ jx : By the homogenized operator along the ej , we shall mean the operator L * 2 ˆ jx u0 = Mj R(x, t) ∂ u0 + S(x, t) ∂u0 − L ∂t ∂t2 s

6

∂/∂xi ( Ai (X1 (∇x u0 , x, t), ..., Xs (∇x u0 , x, t), x, t)) ,

i=1

where Mj = mesβ˜j × length(ej ∩ Q). We determine the homogenized operator ˆx = L

I

ˆ jx . L

j=1

+ ¯εµ Let χ(x/εµ) be a function equal to zero in Bεµ \B and to 1 in the domain 1 0 ≤ χ ≤ 1, χ ∈ C , and |∂χ(z)/∂zi | < c1 , where c1 is independent of ε and µ. Let Ψ(x/ε) be a function equal to zero in the ε−neighborhood of ∂G and to 1 outside the 2ε−neighborhood of ∂G, |∂Ψ(ξ)/∂ξi | < c1 , 0 ≤ Ψ ≤ 1, Ψ ∈ C 1 . We define the vector-valued function ++ Bεµ ,

x x (1) uεµ = u0 + ε(N 1 ( , ∇x u0 , x, t)χ(x/εµ) + N 2 ( , ∇x u0 , x, t)(1−χ(x/εµ)))Ψ(x/ε); ε ε here N 1 (ξ, y, x, t) =

s i=1

(Xie (y, x, t) − yi ) ξi + X0e (y, x, t),

if εξ belongs to the section with the axis segment eε (eε = {x : x/ε ∈ e},) N 1 = 0 outside the sections, and N 2 (ξ, y, x, t) is a function piecewise constant in ξ, equal to

262


s i=1

(Xie (y, x, t) − yi ) ξi0 + X0e (y, x, t)

in the 3c0 µε−neighborhood of the nodal point ξ 0 , which is the end point for e, and equal to zero in the remaining part of Bεµ (1 − χ = 0 on the discontinuity surfaces of N 2 ), and let v be a solution of the homogenized problem ˆ x u0 = L

I j=1

Mj f , x ∈ G, t > 0,

u0 |t=0 = 0,

u0 |∂G = 0

∂u0 |t=0 = 0, ∂t

(we suppose its existence). As above we keep only one of these initial conditions if R = 0, S > 0, or no one of them if R = S = 0. Theorem 5.3.2 Let the solution of the homogenized problem be existing and be having all partial derivatives up to the third order which are bounded by a constant independent on ε, µ. Then the inequality holds: * (1) ∂ 2 (uεµ − uεµ ) + sup sup | ((R ∂t2 t0 ∈[0,T ] ϕ|t=t0 ∈H01 (Bε,µ ∩G) Bε,µ ∩G (1)

S

∂(uεµ − uεµ ) ), ϕ) + (Ai (∇x u(1) εµ , x, t) − ∂t

∂ϕ Ai (∇x uεµ , x, t), ) dx|/(ϕH 1 (Bε,µ ∩G) ∂xi

√ √ C( ε + µ),

6

mes(Bε,µ ∩ G)) |t=t0 ≤

(ε, µ → 0),

where constant C is independent of ε and µ, here ϕ(x, t) is an n−dimensional vector-valued function which belongs to the space H 1 (Bε,µ ∩ G) for each fixed value of t and equal to zero on the boundary ∂G, From this theorem (proved in [16]), one could obtain the generalization of L-convergence and Lt-convergence results of the previous sections to the case of lattice-like domains of general type. Thus to prove theorem 5.2.1 it is sufficient (1) to prove F Lt -convergence of uε,µ − uε,µ to zero.

5.3.3

Proof of Theorem 5.3.2

Proof For every ξ from the nodal domain one obtains: |

∂χ ∂N 1 ∂ |= (N 1 χ + N 2 (1 − χ))| = | χ + (N 1 − N 2 ) ∂ξi ∂ξi ∂ξi

5.3.

263

NON-STATIONARY PROBLEMS |(Xie (y, x, t) − yi )χ + (N 1 − N 2 )

∂χ | ≤ c1 , ∂ξi

with c1 independent of ε and µ; here χ is considered as a function of ξ/µ. Estimate the integral I1 =

B ε,µ∩G

s (1) (1) ∂uεµ ∂ 2uεµ (1) , ϕ) + (Ai (∇uεµ , x, t) , ∂ϕ/∂xi ) dx. , ϕ) + (S (R ∂t ∂t2

i=1

We have I1 = I2 + δ1 , where

I2 =

++ Bε,µ ∩G

s (1) (1) ∂ 2 uεµ ∂uεµ (1) , ϕ) + (Ai (∇uεµ , x, t) , ∂ϕ/∂xi ) dx, (R , ϕ) + (S ∂t ∂t2 i=1

++ |δ1 | = O ϕH 1 (Bε,µ ∩G) mes((Bε,µ \Bε,µ ) ∩ G) = √ = O( µ)ϕH 1 (Bε,µ ∩G)

mes(Bε,µ ∩ G).

The boundedness of the derivatives ∂ l Ni /∂tl (l, i = 1, 2) implies

s ∂v ∂2v (1) (Ai (∇uεµ , x, t) , ∂ϕ/∂xi ) dx + , ϕ) + I2 = (R 2 , ϕ) + (S ++ ∂t ∂t Bε,µ ∩G i=1

+ O(ε)ϕH 1 (Bε,µ ∩G)

Denote I3 =

++ Bε,µ ∩G

s

mes(Bε,µ ∩ G).

(Ai (∇u(1) εµ , x, t) , ∂ϕ/∂xi ) dx .

i=1

Considering the smooth dependency of the functions Ai on its arguments, we have s I3 = (Ai (∇x v + ∇ξ (N 1 Ψ), x, t) , ∂ϕ/∂xi ) dx + ++ Bε,µ ∩G

i=1

+ O(ε)ϕH 1 (Bε,µ ∩G) From

we get

mes(Bε,µ ∩ G).

∂ ((1 − Ψ)N 1 )2L2 (B ++ ∩G) = O(ε) mes(Bε,µ ∩ G) ε,µ ∂ξi

264


I3 =

++ Bε,µ ∩G

s i=1

(Ai (∇x v + ∇ξ N 1 , x, t) , ∂ϕ/∂xi ) dx +

√ + O( ε)ϕH 1 (Bε,µ ∩G) mes(Bε,µ ∩ G).

As ϕ = 0 on ∂G we obtain integrating by parts: I3 = −

−

++ Bε,µ ∩G

++ ∂Bε,µ ∩G

s

(

i=1

s i=1

∂¯ Ai (∇x v + ∇ξ N 1 , x, t) , ϕ) dx − ∂xi

(ni Ai (∇x v + ∇ξ N 1 , x, t) , ϕ) ds +


¯

∂ stands for the full derivative of a function depending on x, t and ξ = Here ∂x i ++ . x/ε, n is an inside normal vector for Bε,µ e Due to property 1) of the Xi , the last integral is

I4 =

=

++ ∂Bε,µ ∩G

x0 : Πx0 ∩G=∅

=

ˆ x0 : Πx0 ⊂G

s i=1

(ni Ai (∇x v + ∇ξ N 1 , x, t) , ϕ) ds = s

++ ∂ Πx0 ∩∂Bεµ ∩G

++ ∂ Πx0 ∩∂Bεµ ∩G

i=1

s i=1

(ni Ai (∇x v + ∇ξ N 1 , x, t) , ϕ) ds = (ni Ai (∇x v + ∇ξ N 1 , x, t) , ϕ) ds +


where the summation is made with respect to all nodal points x0 , such that the ++ which contains x0 , belongs to connected component Πx0 of the set Bεµ \Bεµ ¯ G. Here n is an outside normal of the domain Πx0 (and respectively an inside ++ normal vector for Bε,µ ). 0 For each node x , we set ϕdx Πx0 , ϕΠx0 = mes(Πx0 )

then the following equality is valid: ϕ = ϕΠx0 + (ϕ − ϕΠx0 ), and from lemma 4.A1.2 we obtain the estimate ϕ − ϕΠx0 L2 (∂Π

++ x0 ∩∂Bε,µ )

√ ≤ C εµ∇ϕL2 (Πx0 ) ,

5.3.

265


where C does not depend on ε, µ. Outside G the function ϕ is defined on Bε,µ as zero. Therefore

=

++ ∂ Πx0 ∩∂Bεµ

++ ∂ Πx0 ∩∂Bεµ ∩G

s

(ni Ai (∇x v + ∇ξ N 1 , x, t) , ϕ) ds =

i=1

s

(ni Ai (∇x v|x=x0 + ∇ξ N 1 |x=x0 , x0 , t) , ϕ) ds +

i=1

+ O(εµ) =

s

++ ∂ Πx0 ∩∂Bεµ

i=1

where |δx0 | ≤ c00

++ ∂ Πx0 ∩∂Bεµ

|ϕ|ds =

(ni Ai (∇x v|x=x0 + ∇ξ N 1 |x=x0 , x0 , t) , ϕΠx0 ) ds + δx0 ,

++ ∂ Πx0 ∩∂Bεµ

|ϕ − ϕΠx0 |ds + εµ

++ ∂ Πx0 ∩∂Bεµ

|ϕ|ds ,

with the constant c00 being independent of ε, µ, x0 and ϕ. Property 2) of Xi implies that the integral

++ ∂ Πx0 ∩∂Bεµ ∩G

s i=1

(ni Ai (∇x v|x=x0 + ∇ξ N 1 |x=x0 , x0 , t) , ϕΠx0 ) ds

vanishes. Estimate now δx0 . Applying the Cauchy-Bunyakowskii-Schwartz inequality and then Lemma 4.A1.2, we get: ≤

++ ∂ Πx0 ∩∂Bεµ

|ϕ − ϕΠx0 |ds ≤

++ ++ mes(∂Πx0 ∩ ∂Bεµ )ϕ − ϕΠx0 L2 (∂Π 0 ∩∂Bεµ ) ≤ x √ ++ )∇ϕL2 (Πx0 ) ≤ ≤ c01 εµ mes(∂Πx0 ∩ ∂Bεµ √ s s ≤ c02 ε µ ∇ϕL2 (Πx0 ) ,

with the constants c01 , c02 being independent of ε, µ, x0 and ϕ. Applying the Cauchy-Bunyakowski-Schwarz inequality and then Lemma 4.A1.1, we get: |ϕ|ds ≤

++ ∂ Πx0 ∩∂Bεµ

++ ∂ Πx0 ∩∂Bεµ

|ϕΠx0 |ds +

266

CHAPTER 5. LATTICE STRUCTURES +

|ϕ − ϕΠx0 |ds.

++ ∂ Πx0 ∩∂Bεµ

Estimate both of these integrals. We get 1 1 ϕL2 (Πx0 ) ≤ | ϕdx| ≤ √ |ϕΠx0 | = mesΠx0 Πx0 mesΠx0

≤ c02

ϕL2 (Πx0 ) ++ εµmes(∂Πx0 ∩ ∂Bεµ )

,

where c02 is a positive constant being independent of ε, µ, x0 and ϕ, such that the following inequality holds true: 1 ++ εµmes(∂Πx0 ∩ ∂Bεµ ). c202

mesΠx0 ≥ So, the integral

++ ∂ Πx0 ∩∂Bεµ

is estimated by

√

s

c02 √ ϕL2 (Πx0 ) εµ

s

|ϕΠx0 |ds

++ mes(∂Πx0 ∩ ∂Bεµ )

that is O( εεµµµ ϕH 1 (Πx0 ) ). As mentioned above, the same estimate holds true for |ϕ − ϕΠx0 |ds ++ ∂ Πx0 ∩∂Bεµ

and hence for

++ ∂ Πx0 ∩∂Bεµ

|ϕ|ds.

√ So, |δx0 | ≤ c03 εs µs ϕH 1 (Πx0 ) , where c03 is a positive constant being independent of ε, µ, x0 and ϕ. From the Cauchy-Bunyakovskii-Schwartz inequality for sums we obtain then, that εs µs ϕ2H 1 (Π 0 ) ≤ |δx0 | ≤ c04 s x ε 0 0 x : Πx0 ∩G=∅

x : Πx0 ∩G=∅

Thus,

√ ≤ c05 µ mes(Bε,µ ∩ G)ϕH 1 (Bε,µ ∩G) .

√ √ I4 = O( µ + ε)

mes(Bε,µ ∩ G)ϕH 1 (Bε,µ ∩G) ,

5.3.

267


and

I2 =

++ Bε,µ ∩G

s ∂¯ ∂v ∂2v 1 Ai (∇x v+∇ξ N , x, t) , ϕ) dx + ( , ϕ) − (R 2 , ϕ) + (S ∂xi ∂t ∂t i=1

√ √ + O( ε + µ)ϕH 1 (Bε,µ ∩G) mes(Bε,µ ∩ G).

We denote by Qk1 ...ks the cube {x ∈ IRs |εki ≤ xi ≤ ε(ki + 1)}. The number of cubes, such that Qk1 ...ks ∩ Bε,µ ∩ ∂G = ∅, does not exceed constε1−s and consequently the total measure of these sets is O(ε) mes (Bε,µ ∩G); therefore I5 = −

= −

++ Bε,µ ∩G

k1 ...ks :Qk1 ...ks ∩Bε,µ ⊂G

= −

(

i=1

∂¯ Ai (∇x v + ∇ξ N 1 , x, t) , ϕ) dx = ∂xi s

++ Qk1 ...ks ∩Bε,µ

(

i=1

∂¯ Ai (∇x v+∇ξ N 1 , x, t) , ϕ) dx + ∂xi

√ + O( ε)ϕH 1 (Bε,µ ∩G) mes(Bε,µ ∩ G) =

k1 ...ks :Qk1 ...ks ∩Bε,µ ⊂G

|xi =ε(ki +1/2),

s

i=1,...,s

s

++ Qk1 ...ks ∩Bε,µ

i=1

(

∂¯ Ai (∇x v+∇ξ N 1 , x, t) , ϕ) ∂xi

√ dx + O( ε)ϕH 1 (Bε,µ ∩G) mes(Bε,µ ∩ G).

++ Next, by using on each set Qk1 ...ks ∩ Bε,µ the representation

ϕk1 ...ks

ϕ = ϕk1 ...ks + (ϕ − ϕk1 ...ks ), 1 = ϕ(x, t) dx, mes(Qk1 ...ks ∩ Bε,µ ) Qk1 ...ks ∩Bε,µ

´ inequality of Appendix 4.A2 the Poincare-Friedrichs ϕ − ϕk1 ...ks L2 (Qk1 ...ks ∩Bε,µ ) ≤ c05 ε∇ϕL2 (Qk1 ...ks ∩Bε,µ ) , c05 does not depend on ε and µ), the Cauchy - Bunyakovskii - Schwartz and J J J i=1 a2i , we get inequality for sums: i=1 ai ≤ |ϕ− < ϕ > | dx = k1 ...ks :Qk1 ...ks ∩Bε,µ ⊂G

= O ε−s

++ Qk1 ...ks ∩Bε,µ

k1 ...ks :Qk1 ...ks ∩Bε,µ ⊂G

(

++ Qk1 ...ks ∩Bε,µ

|ϕ− < ϕ > | dx)2

=

268


= O

,

ε−s εs µs−1

k1 ...ks :Qk1 ...ks ∩Bε,µ ⊂G

+ O(ε)ϕH 1 (Bε,µ ∩G)

I5 = −

k1 ...ks :Qk1 ...ks ∩Bε,µ ⊂G

ϕ− < ϕ >

2L2 (Q ++ k1 ...ks ∩Bε,µ )

=

mes(Bε,µ ∩ G).

++ Qk1 ...ks ∩Bε,µ

∂2v 1 ˆ Lx v − R 2 − M ∂t

√ √ ∂v ϕk1 ...ks dx + O( ε + µ)ϕH 1 (Bε,µ ∩G) mes(Bε,µ ∩ G). −S ∂t

From presentation ϕ = ϕk1 ...ks + (ϕ − ϕk1 ...ks ), we have again

−

s

++ Bε,µ ∩G i=1

−

Bε,µ ∩G

(

∂¯ Ai (∇x v + ∇ξ N 1 , x, t) , ϕ) dx = ∂xi

∂v ∂2v 1 ˆ ( Lx v − R 2 − S ), ϕ dx + ∂t ∂t M


This implies finally

I1 =

Bε,µ ∩G

=

Bε,µ ∩G

√ √ f (x, t)ϕdx + O( ε + µ)ϕH 1 (Bε,µ ∩G) mes(Bε,µ ∩ G) =

∂uεµ ∂ 2 uεµ ), ϕ +S (R ∂t ∂t2

+

s

(Ai (∇x uεµ , x, t) ,

i=1


Theorem is proved. (1) Thus the sequence uεµ − uεµ FLt-converges to zero .

∂ϕ ) dx + ∂xi

5.4.

5.4

269

L- AND FL-CONVERGENCE IN ELASTICITY

L- and FL-Convergence in elasticity

Consider the elasticity system of equations s ∂uεµ ∂ ) = f (x), (Aij ∂xj ∂xi i,j=1

x ∈ Bε,µ ⊂ IRs ,

with the boundary condition s

ni (Aij

i,j=1

∂uεµ ) = 0, ∂xj

x ∈ ∂Bε,µ ,

where Aij are constant s × s−matrix-valued functions with the elements akl ij , satisfying the conditions kj lk 1) akl ij = ail = aji ∀ k, i, l, j ∈ {1, ..., s} and 2) there exists a constant C0 > 0 such that s

i,j,k,l=1

akl ij ηlj ηki ≥ C0

s

2 ηlj

l,j=1

for each symmetric constant matrix (ηlj )1≤l,j≤s . In this case the estimation of Lemma 5.2.2 is not valid because of the dependence of the constant of Korn’s inequality of µ. Indeed the following lemma takes place: Lemma 5.4.1 (proved in [145]).Let

EBε,µ ∩G (ϕ) =

s

Bε,µ ∩G i,j=1

(eji (ϕ))2 dx, eji (ϕ) =

∂ϕj 1 ∂ϕi + ). ( ∂xi 2 ∂xj

Then for any lattice Bε,µ and domain G, independent of ε and µ with piecewise-smooth boundary there exists such vector-valued function vε,µ ∈ H 1 (Bε,µ ∩ G) that vε,µ |Bε,µ ∩∂G = 0, ∇vε,µ 2L2 (Bε,µ ∩G) ≥ c1 µs−1 εs−2 , EBε,µ ∩G (vε,µ ) ≤ c2 µs εs−2 , i.e. EBε,µ ∩G (vε,µ ) ≤ c3 µ∇vε,µ 2L2 (Bε,µ ∩G) , where constants c1 , c2 , c3 do not depend on ε and µ. The proof is based on the analogous proposition for a bar π = (−1, 1) × βµ , where βµ = {(x′ /µ ∈ β}, β is (s − 1)−dimensional bounded domain with a piecewise-smooth boundary (satisfying the cone condition; in the twodimensional case β is an interval), i.e. there exist such constants c1 , c2 , c3 > 0,

270


independent of µ, and the s-dimensional vector-valued function wµ (x), such that wµ ∈ H 1 (π) and wµ |x1 =±1 = 0, wµ 2L2 (π) ≥ c1 µs−1 ,

∇wµ 2L2 (π) ≥ c2 µs−1 ,

Eπ (wµ ) ≤ c3 µs .

This example was constructed in subsection 4.4.1. By the linear application we could build such function wµ , which satisfies to all conditions of the proposition and vanishes in the vicinity of the planes {x1 = ±1}. If now for each bar of the lattice structure Bε,µ except of those that cross the boundary of the domain G we shall construct the corresponding function (with the replacement of µ by the product εµ and of the length 2 of the bar by ε) we obtain the proof of this lemma. Thus L-convergence for the elasticity equations is not the consequence of the F L−convergence. Nevertheless the Korn inequality for the lattice structures with the constant estimated by a power µ−K with some K was proved in [145] and then this constant was calculated more precisely in [129], [43] and then in [68]. The Korn inequality for lattices was used with boundary layer and homogenization methods for the construction of the asymptotic expansion of the solution of the elasticity system of equations, set in a lattice-like domain ( [148]). Here above we have introduced the L-convergence and the FL-convergence for the lattice-like domains and we have applied it to the homogenization of linear and non-linear problems in these domains. The techniques of proof of L-convergence is quite different of the techniques of extension of the solution of the problems for perforated domain inside the perforations [45]. More detailed information about the solution could be obtained from the asymptotic expansion of the solution, built in [136], [148]. Using the terminology of Ph.Ciarlet [38], one can say that lattice-like domains are multi-structures with a great number of junctions. Particularly the case of junction of two elastic rods was considered in [94], [95]. The extension techniques was developed for lattice-like structures in [46],[44], where the convergence results were obtained for the elasticity system of equations, set in these structures. The influence of the order of the passage to limit by small parameters was studied. The boundary value problems considered above for lattice-like domains were of Neumann’s type on ∂Bε,µ . The Dirichlet’s problems for lattice-like domains are developed in [90] (periodic case) and [91] (non-periodic case).

5.5

Conductivity of a net

Consider the limit problem of section 4.4 on a periodic net. Define net B as a connected periodic in all xi (i = 1, . . . , s) with period 1 union of segments in Rs

271

5.5. CONDUCTIVITY OF A NET : B=

+∞ 9

en

n=−∞

such that (i) any two segments ei , ej (i = j) may have only one common point, which is an end-point for both segments. The end-points of the segments are called nodes. (ii) the periodicity cube Q = {x ∈ Rs | − 12 ≤ xi ≤ 12 } intersects finitely many segments e1 , . . . , eM from B ; we assume that B ∩ Q is connected.

Let Bε = {x ∈ Rs | xε ∈ B}. For any segment ei ⊂ B, denote eεi = {x ∈ Rs | xε ∈ ei }. Introduce local longitudinal variable for any eεi in a following way.

Figure 5.5.1. A net. Fix an end point x0 and define for any x ∈ eεi the variable x 1 (x) =

s i=1

γi (xi −

x0i ) where (γ1 , . . . , γs ) are the directing cosines of the vector eεi with the origin at x0 ; so x 1 (x) is the first component of the vector α(x − x0 ) ; here α is an such orthogonal matrix ; α∗ is the matrix of orthogonal space transformation Π → − that the image of the vector i parallel to the x1 axis of length |eεi | is the vector eεi . Consider the problem : d2 uε = f (x), d x21

mes βe

e(x0 )

in the nodes x0 , ueε1 = ueε2

x ∈ eεi , dueε = 0, dx 1 ∀e1 , e2

(5.5.1) (5.5.2)

(5.5.3)

272


with the common node x0 . Here x 1 is defined once for any segment eεi (i.e. we fix direction once) ; mesβe are some positive numbers, associated with any segment e, 1 periodic on Bε . Suppose that f is a T −periodic function of x such that f ∈ C ∞ (Rs ), and that f (x)dx = 0, f (x)d x1 = 0, QT ∩Bε

QT

where QT = (0, T )s , T is a multiple of ε. We seek a T −periodic solution uε ( x1 (x)) on Bε ; T is a multiple of ε. The asymptotic expansion is sought in a form of series (∞) uε

∼

∞ l=0

ε

l

|i|=l

Ni1 ...il

x 1 (x) ε

∂lv . Applying the Bakhvalov’s pro∂xi1 . . . ∂xis

cedure described above in Chapter 2 we obtain equations on Ni1 ...il :

dNi2 ...il dN d2 Ni1 ...il + 2γi1 + γi1 γi2 Ni3 ...il = hi1 ...il , ξ ∈ e ⊂ B, dξ2 dξ1

(5.5.4)

1

mes βe

e(ξ0 )

d Ni1 ...il + γi1 Ni2 ...il dξ1

= 0,

(5.5.5)

in the nodes ξ0 of B (the direction of ξ1 is chosen and fixed once) ; Ni1 ...il is continuous and 1 periodic on B d Ni2 ...il + γie2 Ni3 ...il dξ1 / mes βe |e|; mes βe γie1 hi1 ...il = dξ1 ecQ∩B ecQ1 ∩B1 e (5.5.6) for l = 1 we get d d Ni1 + γi1 = 0 , ξ ∈ e ⊂ B, (5.5.7) dξ1 dξ1

e(ξ0 )

mes βe

d Ni + γi1 dξ1 1

= 0 , ξ = ξ0 ,

(5.5.8)

ξ0 is a node of B. It can be proved that the matrix (hi1 i2 )1≤i1 ,i2 ≤s is symmetric positive definite matrix (see below Lemma 5.7.3). Thus we can organize a recurrent process of determining of coefficients of series V =

∞ j=0

εj Vj (x).

(5.5.9)

273

5.6. ELASTICITY OF A NET We obtain a recurrent chain of T −periodic problems Vj = fj (x) , LV

where

fj (x) = We prove by induction that

= L

j−1

s

hi 1 i 2

i1 ,i2 =1

∂2 ∂xi1 ∂xi2

hi1 ...ij−k Di Vk .

(5.5.10)

(5.5.11)

k=0 |i|=j−k

fj (x)d x = 0 ; so the chain of problems is solv-

QT

able.

Justification in a standard way gives an estimate for the difference of a T −periodic solution of problem (5.5.1)-(5.5.3) with vanishing average and of a partial sum u(K) ε

=

K+1 l=0

ε

l

i1 ...il ∈{1,...,s}

Ni1 ...il

x 1 (x) ε

Di1 ...il

k

εj Vj (x)

(5.5.12)

j=0

with vanishing average : uε(k) − uε H 1 (QT ∩Bε ) = O(εk ).

(5.5.13)

The main term is described by the equation V0 = f (x) LV

(5.5.14)

with T periodicity condition ; is determined by relations (5.5.7),(5.5.8),(5.5.10). here operator L

5.6

Elasticity of a net

Consider the limit problem of section 4.5 on a periodic net Bε in two-dimensional case. It consists of (i) the equations ⎞ ⎛ d2 u ˜1ε e E d˜ x21 ⎠ = f (x), x ∈ eε , (5.6.1) αe∗ ⎝ ˜2 e d4 u −E < ξ22 >β µ2 d˜xε4 1

where

αe =

γ1e −γ2e

γ2e e , γ1

is an orthogonal matrix associated with any segment e. Here β = (−1/2, 1/2), and < ξ22 >β = 1/12 and mes β = 1 with directing cosines (γ1e , γ2e ).

274

CHAPTER 5. LATTICE STRUCTURES The interface conditions are: for any node x0 ,

mes βe γ e

e(x0 )

and

d˜ u1ε e =0 d˜ x1

mes βe γ e < ξ22 >

e(x0 )

d˜ u2

(5.6.2)

d2 u ˜2ε e = 0; d˜ x21

(5.6.3)

e

˜2ε = 0 in all nodes, moreover, γ e u ˜1ε , d˜xε1 are continuous functions on Bε and u e e e ∗ γ = (γ1 , γ2 ) . Here E is the Young modulus, f ∈ C ∞ is a T −periodic s−dimensional vector-valued function, T is a multiple of ε. We assume as well that the following integrals of the right hand-side vanish: f (x)dx = 0, f (x)dx = 0. (0,T )2 ∩Bε

(0,T )2

We seek the solution u(x) in the class of T −periodic two-dimensional vectorvalued functions. The asymptotic expansion is sought in a form of series u(∞) ε

∼

∞ l=0

ε

l

|i|=l

Ni1 ,...,il

x ˜1 (x) ε

∂ l V (x) , ∂xi1 . . . ∂xil

(5.6.4)

where Ni (ξ) are 2 × 2 1−periodic in ξ matrices, V is a two-dimensional vector, ˜ e (ξ)αe , Ni (ξ) = αe∗ N i N∅ (ξ) = I, ˜ e (ξ) = N ˜e N i i

(1)

˜e (ξ) + µ−2 N i

(2)

(ξ),

(5.6.5)

˜ e (1) (ξ) has a vanishing second line; where I stands for the identity matrix, N i e (2) ˜ N (ξ) has a vanishing first line. i Applying the Bakhvalov procedure (section 2.2) we obtain the equations on ˜ e (1) : ˜ e (1) )1 of N the first component (N i1 ,...,il i1 ,...,il *

˜ e (1) )1 ˜ e (1) )1 d(N d2 (N i1 ,...,il i2 ,...,il ˜ e (1) )1 + γi1 γi2 (N E + 2γ i 1 i3 ,...,il 2 ˜ ˜ dξ1 dξ 1

6

= (αe hi1 ...il αe∗ )1 (5.6.6)

for any e,

275

5.6. ELASTICITY OF A NET

E mes βe γe

e(ξ0 )

*

e (1)

˜ d(N i1 ,...,il )1 ˜ e (1) )1 + γi1 (N i2 ,...,il dξ˜1

6

=0

(5.6.7)

in the nodes ξ0 ; e (1) ˜ e (1) (ξ)αe , are 1-periodic and continuous functions; Ni = αe∗ N i e (2) ˜e we obtain as well the following chain of problems for Ni = αe∗ N i

−E
βe

*

(2)

(ξ)αe :

˜ e (2) )2 ˜ e (2) )2 ˜ e (2) )2 d2 (N d3 (N d4 (N i3 ,...,il i2 ,...,il i1 ,...,il + 6γ γ + 4γ i i i 1 2 1 3 4 ˜ ˜ dξ dξ˜2 dξ 1

1

1

e (2)

˜ d(N i4 ,...,il )2 ˜ e (2) )2 + γi1 γi2 γi3 γi4 (N +4γi1 γi2 γi3 i5 ,...,il dξ˜1

6

= (αe hi1 ...il αe∗ )2

(5.6.8)

for any e,

E
βe mes βe

e(ξ0 )

*

e (2)

e (2)

˜ ˜ d2 (N d(N i1 ,...,il )2 i2 ,...,il )2 ˜ e (2) )2 +γi1 γi2 (N +2γi1 i3 ,...,il 2 ˜ dξ˜1 dξ 1

6

=0

(5.6.9)

in the nodes ξ0 ; e (2) )2 1 ,...,il

˜ d(N i

dξ˜1 ˜ e (2) is N i

˜ e (2) )2 is continuous on B; + γi1 (N i2 ,...,il

a 1-periodic function vanishing in all nodes ξ0 ; hi1 ,...,il =

e⊂Q1 ∩B

E mes

e

˜ e (1) )1 +γi1 γi2 (N i2 ,...,il

6

βe αe∗

*

γi1

˜ αe dξ1 /

e⊂Q1 ∩B

e (1)

˜ d(N i1 ,...,il )1 + dξ˜1 mes βe |e|;

(5.6.10)

here N∅e = I (identity matrix) and for any square matrix M, (M )k is the k−th line of M. ˜ e (1) and N ˜ e (2) are proved in section 5.8. The solvability of problems for N i i 2 i 1 i2 ˆ= is an operator of elasticity. Assume that the operator L i1 ,i2 =1 hi1 i2 D Then we organize a recurrent procedure of determining of coefficients of series V =

∞ j=0

as in the previous section. The main term is

εj Vj (x)

(5.6.11)

276


2 ∂ 2 V0 ˜1 (x) ε 2 e (2) x ; Ni1 ,i2 V0 + ( ) ∂xi1 ∂xi2 ε µ i ,i =1 1

(5.6.12)

2

here V0 is a T −periodic solution of homogenized equation

ˆ V0 = f (x), LV (5.6.13) ˆ are determined as in [16] : where the coefficients hi1 i2 of the operator L * ˜ e (1) ) d(N i2 ∗ hi 1 i 2 = E mes βe αe γi1 + dξ˜1 e e⊂Q1 ∩B

˜e +γi1 γi2 (N ∅

(1)

here ˜e N ∅

(1)

6

˜ ) αe dξ1 /

= (αe αe∗ )1 =

i.e.

γi1 γi2

e⊂Q1 ∩B

e

(γ1e )2 −γ1e γ2e

mes βe |e|,

e⊂Q1 ∩B

0 , 0

1 0

hi 1 i2 = * ˜ e (1) )1 d(N γ1e i2 αe + E mes βe γ i1 γ2e dξ˜1 6 γ1e γ2e dξ˜1 / mes βe |e|, (γ2e )2

(5.6.14)

e⊂Q1 ∩B

˜ e (1) αe is a 1-periodic linear for any e solution to the problem where N i1 * 6 ˜ e (1) ) d(N 1 0 i1 E mes βe γe αe + γi1 αe = 0 (5.6.15) 0 0 dξ˜1 e e⊂Q1 ∩B

e (1)

˜ in the nodes ξ0 such, that αe∗ N αe is a continuous function; i1 e (2) ˜ N αe is a 1-periodic solution to the problem i 1 i2

E < ξ22 >βe

˜ e (2) )2 d4 (N i1 i2 = (αe hi1 i2 αe∗ )2 , ξ ∈ e, dξ˜4

(5.6.16)

1

E mes βe
βe

˜ e (2) ) d2 (N i 1 i2 αe dξ˜2 1

= 0, ξ = ξ0 , e (2)

(5.6.17)

i1 i2 ˜ and is a continuous on B function; αe∗ N i1 i2 αe is a 1-periodic on B dξ˜1 vanishing at all nodes matrix-valued function.

277

5.6. ELASTICITY OF A NET

In particular for the net B = B(1) ∪ B(2) with B(i) = ∪+∞ k=−∞ {ξ = (ξ1 , ξ2 )|ξi ∈ IR, ξ3−i = k}, here mes β = 1. We get (cf [16]) ˆ = Lu e (1)

and Ni1

E 2 E 2

∂ 2 u1 ∂x21 ∂ 2 u2 ∂x22

= 0; hi1 i2 = δi1 i2 E2 h( i1 +i2 ) and h(1) = 2

e (2)

and Ni1 i2 = 0 for i1 = i2 ;

(5.6.18)

1 0 0 , h(2) = 0 0 0

0 , 1

˜ e (2) = N 11

5

(0, 0) on B(1) (0, φ(ξ˜1 )) on B(2) ∩ {ξ2 ∈ (0, 1)}

(5.6.19)

˜ e (2) = N 22

5

(0, 0) on B(2) (0, φ(ξ˜1 )) on B(1) ∩ {ξ1 ∈ (0, 1)}

(5.6.20)

and

Here φ(ξ˜1 ) = − 14 ξ˜12 (1 − ξ˜1 )2 . So, 5 (2)

N11 =

and (2) N22

=

5

(0, 0) on B(1) (φ(ξ2 ), 0) on B(2) ∩ {ξ2 ∈ (0, 1)}

(5.6.21)

(0, φ(ξ1 )) on B(1) ∩ {ξ2 ∈ (0, 1)} (0, 0) on B(2)

(5.6.22)

(2)

(2)

So, we get < N22 >β = −(0, 1/240), < N11 >β = (1/240, 0). Thus,

0

∂ 2 V0 ˜1 (x) ε 2 e (2) x = V0 + ( ) Ni1 ,i2 ) ∂xi1 ∂xi2 ε µ

/

β

ε V0 + ( )2 (1/240) µ

1 0 ∂ 2 V0 − 0 0 ∂x21

0 0

0 ∂ 2 V0 = 1 ∂x22

∗ ε 2 ∂ 2 V01 ∂ 2 V02 = , V0 + (1/240)( ) ∂x22 ∂x21 µ ε V0 + (1/240)( )2 (2/E)f (x) = µ

V0 +

ε 1 ( )2 f (x). 120E µ

(5.6.23)

We obtained thus the average of the leading term of the asymptotic in twodimensional case. It satisfies the following homogenized equation:

278


∗ 2 2 ∂ f ∂ f ε ε 1 2 1 2 2 ˆ V0 + . , ( ) f (x)) = f (x) − (1/240)( ) L(V ∂x21 ∂x22 µ 120E µ

(5.6.24)

So, the average of the main term is a solution to the homogenized equation with the operator obtained by principle of splitting ( [134],[16]) with an additional term in the right hand side. The similar effect was obtained recently in [67] for a ”regularized” elasticity equation, containing the term proportional to the displacement. The important difference of the elasticity equation in comparison to the conductivity equation is that there is a corrector in (5.6.12) of order ( µε )2 . It can be great or of order of 1. This corrector can be easily explained ”mechanically”: its presence is due to the orthogonal (with respect to every segment e) components of the force f (x).

5.7

Conductivity of a lattice: an expansion

We consider equation Lx u = ∆u = f (x), x ∈ Bµ,ε ,

(5.7.1)

with boundary condition ∂u = 0, x ∈ ∂Bµ,ε , ∂n

(5.7.2)

where f ∈ C ∞ is a T −periodic function, T is a multiple of ε and as above s = 2 or s = 3. We seek the solution u(x) in the class of T −periodic functions of HT1 −per (Bµ,ε ). Here HT1 −per (Bµ,ε ) stands for the completion of the set of T −periodic differentiable functions defined at Bµ,ε with respect to the H 1 (Bµ,ε ∩ (0, T )s )−norm. Solution to this problem is defined as a function uµ,ε ∈ HT1 −per (Bµ,ε ), such that

∀v ∈

HT1 −per (Bµ,ε ),

−

Bµ,ε ∩(0,T )s

(∇uµ,ε , ∇v) dx =

f (x) v dx.

Bµ,ε ∩(0,T )s

Below we construct the asymptotic expansion of the solution to problem (5.7.1)-(5.7.2) as µ, ε → 0 under the assumption that the following integrals of the right hand-side vanish: f (x)dx = 0, f (x)dx = 0. (0,T )s

Bµ,ε ∩(0,T )s

Problem (5.7.1)-(5.7.2) simulates the conductivity of a frame structure (or lattice structure). A geometric frame model depending on two small parameters was first considered in [56] (it was called there ”skeletal structure”.)

5.7.

279

CONDUCTIVITY OF A LATTICE: AN EXPANSION

Existence of solution is a well known corollary of the Riesz theorem on presentation of a linear bounded functional in a Hilbert space (or the LaxMilgramm lemma) and the following Poincar´ é inequality: Theorem 5.7.1 There exist a constant CP > 0, such that for all sufficiently small ε and µ, for all u ∈ HT1 −per (Bµ,ε ), u2L2 (Bµ,ε ∩(0,T )s )

≤

+

1

mes Bµ,ε ∩ (0, T )s

CP εs−1

u dx

Bµ,ε ∩(0,T )s

2

+

(∇ u)2 dx.

Bµ,ε

∩(0,T )s

Proof is given in the Appendix to this Chapter. Theorem 5.7.2For all sufficiently small ε and µ, for all strictly positive α, for all f0 , f1 , ..., fs ∈ L2 (Bµε ∩ (−α, α)s ), T −periodic in x1 , ..., xs , the problem ∆u = f0 (x) +

with boundary condition

s ∂ fi (x), x ∈ Bµ,ε , ∂x i i=1

s

∂u = ni fi (x), x ∈ ∂Bµ,ε , ∂n i=1 has a T −periodic solution in HT1 −per (Bµ,ε ) if and only if f0 (x)dx = 0. Bµ,ε ∩(0,T )s

There exist a constant independent of µ, ε, such that ∇uL2 (Bµ,ε ∩(0,T )s ) ≤

s C

εs−1

i=1

ffi L2 (Bµ,ε ∩(0,T )s ) .

Proof. As it was mentioned above, the existence of solution is a well known corollary of the Riesz theorem on presentation of a linear bounded functional in a Hilbert space (or the Lax-Milgramm lemma). To this end we consider the subspace u(x)dx = 0}. The of HT1 −per (Bµ,ε ) that is {u ∈ HT1 −per (Bµ,ε ), Bµ,ε ∩(0,T )s variational formulation with a test function equal to a solution to the problem gives the estimate

∇u2L2 (Bµ,ε ∩(0,T )s )

≤

s i=1

ffi L2 (Bµ,ε ∩(0,T )s ) ∇uL2 (Bµ,ε ∩(0,T )s ) +

280

CHAPTER 5. LATTICE STRUCTURES +ff0 L2 (Bµ,ε ∩(0,T )s ) uL2 (Bµ,ε ∩(0,T )s ) .

Applying now the Poincar´ é inequality of Theorem 5.7.1, we obtain the estimate of Theorem 5.7.2. Let us describe the asymptotic procedure of construction of solution to problem (5.7.1)-(5.7.2). At the first step for every segment e an asymptotic solution is sought in the form similar to that of section 4.3 and at the second step the boundary layer correctors are constructed in the same way as in section 4.3. Functions ve set on the net satisfy the equations and the junction conditions analogous to that of section 5.5, therefore its asymptotic solution is sought in the form of series (∞) similar to uε of section 5.5 (i.e. in the form of (5.7.7)). Of course this form is taken into account in the Taylor expansion procedure for the ansatz in the neighborhoods of the nodes (5.7.15). For the macroscopic unknown function V of the expansion (5.7.7) we obtain the homogenized equation similar to that of section 5.5. (according to the principle of splitting of the homogenized operator [ 134], [16]). As usual, function V is expanded in the regular series (see (5.7.36)) and we obtain finally the recurrent chain of problems for the terms of this series. The justification is a standard procedure of the truncation of the expansion, its substitution into the equation and the boundary conditions and the estimation of the discrepancy. Then we apply the a priori estimate and get the estimate for the difference of the exact solution and the truncated expansion. 1. So, first , we construct the asymptotic corresponding to a segment e = enh in B . We change to new variables x ˜ = α(x − εh). In these new variables, the ˜ α } becomes the rod. cylinder {x/ε ∈ B hj we set

Uεµ = {˜ x1 ∈ IRs | 0 < x ˜ < lε, x ˜′ /εµ ∈ βj }; Γεµ = {˜ x1 ∈ IRs | 0 < x ˜ < lε, x ˜′ /εµ ∈ ∂β βj }.

In what follows β = βj . We write f˜(˜ x) = f (α∗ x ˜+εh) and u ˜(˜ x) = u(α∗ x ˜+εh) .

˜ α becomes Then problem (5.7.1), (5.7.2) on B hj ∆u ˜ = f˜(˜ x), x ˜ ∈ Uεµ ,

(5.7.3)

∂u ˜ = 0, x ˜ ∈ Γεµ . ∂n

(5.7.4)

We make the substitution ξ ′ = (ξ˜2 , . . . , ξ˜s ), ξ˜i = x ˜i /(εµ) and expand f˜ in an asymptotic series with respect to εµξ˜i , of the form f˜ ∼

∞ j=0

(εµ)j

rj r=1

Fjr (ξ˜2 , . . . , ξ˜s )ψ˜jr (˜ x1 ),

5.7.

281


In this formula the functions ψ˜jr (˜ x1 ) are the values of the derivatives q

s−1 ∂ j f /∂xq11 ...∂xs−1 ∂xs j−q1 −...qs−1

at the point x = α∗ (˜ x, 0, . . . , 0) + εh, Fjr (ξ ′ ) are some polynomials. In particular, ψ˜00 (˜ x1 ) = f˜( ˜1 , 0, ..., 0), and F00 = 1 and r0 = 1. We represent Fjr in the form F¯jr + F˜jr , where F˜jr β = 0 and F¯jr = F Fjr β . Using superposition , we seek the solution to problem (5.7.3) , (5.7.4) in the form : u ˜∼

(∞)

(εµ)j

rj

u ˜jr ,

(5.7.5)

r=1

j=0

where u ˜jr is the solution of the problem for a right hand side of the form Fjr ψ˜jr . From now on in this section we omit the ∼ that denotes the change of variables, fix j and r and do not mention them explicitly when this does not create ambiguity . We seek a formal asymptotic solution to problem (5.7.3)-(5.7.4) with right hand side Fjr ψjr in the form of a series

u ˜∞ ˜ ( ˜1 ) + (εµ)2 jr ∼ ω

∞

(εµ)l Ml

l=0

x x1 ) ˜′ dl ψ˜jr (˜ l εµ d˜ x1

(5.7.6)

where Ml are the functions of Remark of section 2.3; ω ˜ (˜ x1 (x)) = vεµ (x) + ε2 wεµ (x), vεµ (x) and wεµ (x) are defined on IRs , and

vεµ (x) ∼ V (x) +

∞ ∞

m=1 q=0

wεµ (x) ∼

εm µq

∞ ∞

m=0 q=0

(i1 ...im ): in ∈{1...s}

εm µq

∂ m V (x) x˜1 (x) ˜q ; ) N ( i1 ...im ∂xi1 . . . ∂xim ε

(j1 ...js ): j1 +...+js =j (i1 ...im ): in ∈{1...s}

∂j f ∂m x˜1 (x) ˜q . ) ×M i1 ...im ,j1 ...js ( ∂xi1 . . . ∂xim ∂xj11 . . . ∂xjss ε

(5.7.7)

˜ ˜ ˜q ˜q Here V (x) is a smooth T −periodic function, N i1 ...im (ξ1 ) and Mi1 ...im ,j1 ...js (ξ1 ) are functions defined on every segment e. We note that ψ˜jr can be represented in the form ψ˜jr ( ˜1 (x)) =

(j1 ...js ): j1 +...+js =j

∂ j f (x) ˜ j ...j ( x ) , K 1 s j ε ∂x1 1 . . . ∂xs js

282


˜ j ...j are constant on e . where K 1 s The substitution of (5.7.6) in (5.7.3) and (5.7.4) is described in section 2.3 (with µ replaced by εµ ). As a result, we find that (5.4.4) is satisfied asymptotically exactly and that (5.7.3) yields ”the homogenized over the rod” equation, of the form ∞ l˜ d2 ω ˜ l M d ψjr ˜jr + ∼ 0. F (εµ) h ψ − jr l d˜ x21 d˜ xl1

(5.7.9)

l=0

Here we have not yet used the fact that ω has the form (5.7.7). We substitute now (5.7.7) in (5.7.9) and gather together the coefficients of the same powers of εµ and the derivatives Di V =

∂mV ∂j f . , Dj f = j 1 ∂xi1 . . . ∂xim ∂x1 . . . ∂xjss

We obtain an asymptotic equality of the form ∞ ∞ x ˜1 (x) ∂mV q εm−2 µq HiN1 ...i m ∂xi1 . . . ∂xim ε m=1 q=0 (i1 ...im )

+

(j1 ...js ):j1 +...+js =j

HiM1 ...iq m , j1 ...js

x ∂j f ∂m ˜1 (x) ∂xi1 . . . ∂xim ∂xj11 . . . ∂xjss ε

∼ 0,

(5.7.10)

q q ˜ ˜q HiN1 ...i = RN i1 ...im (ξ1 ) + Ti1 ...im (N ), m

HiM1 ...iq m ,

j1 ...js

˜q = RM i1 ...im ,

Tiq1 ...im (N )(T Tiq1 ...im ,

j1 ...js (M))

˜ +Tq i1 ...im ,

j1 ...js (ξ1 )

j1 ...js (M),

(5.7.11)

are linear combinations of the

˜q Niq1 ...im (M i1 ...im ,

j1 ...js )

and their derivatives with multi-index (i1 . . . im ) of length m1 less than m or m1 = m with superscript q1 < q, and R is a differential operator along the variable ξ˜1 = x ˜1 (x)/ε, of the form R =

∂2 . ∂ ξ˜12

q Ñ ˜M We equate H j1 ...js (ξ1 ) and Hi1 ...is , j1 ...js (ξ1 ) with the constants hi1 ...im and We obtain a homogenized equation for V of the form:

hqi1 ...im , j1 ...js .

5.7.

∞ ∞

ε

m−2 q

µ

m=1 q=0

+

283


hqi1 ...im

(i1 ...im )

hqi1 ...im , j1 ...js

(i1 ...im ) (j1 ...js ): j1 +...+js =j

∂mV ∂xi1 . . . ∂xim

∂j f ∂m j 1 ∂xi1 . . . ∂xim ∂x1 . . . ∂xjss

∼ 0. (5.7.12)

The coefficients hqi1 ...im , and hqi1 ...im , j1 ...js of this equation will be determined later. ˜q ˜q Thus the N i1 ...im and Mi1 ...im , j1 ...js are determined from the equations q ˜ q i ...i = T˜ q RN 1 m i1 ...im (N ) + hi1 ...im ;

˜ q i ...i , RM 1 m

js ...js

=

T˜iq1 ...im , js ...js (M )

+

(5.7.13)

hqi1 ...im

(5.7.14)

and some interface conditions formulated below. (∞) (∞) Let u ˜ε = u ˜jr (x) the asymptotic series (5.7.6), (5.7.7), which is the solu˜ α }. tion of (5.7.1), (5.7.2) in the cylinder {x : xε ∈ B hj We expand the functions

∂ l ψjr ∂m ∂j f ∂mV , , , j j 1 s ∂xi1 . . . ∂xim ∂xi1 . . . ∂xim ∂x1 . . . ∂xs ∂x ˜li

˜ i ...i ,j ...j ( x˜1 (x) ) in Taylor series in the neighborhood of ˜i ...i ( x˜1 (x) ) and M N 1 m 1 s 1 m ε ε the point x0 corresponding to the node xε0 , and substitute these expansions in (5.7.6 ) . We obtain

u ˜(∞) ∼ ε

∞ ∞

εl µr

l=0 r=0

(N Nie1 ...irl (

(i1 ...il )

+ε2

∂lV x − x0 (x0 ) ) ∂xi1 . . . ∂xil εµ

Mie1 ...irl ,

j1 ...js (

(j1 ...js ), j1 +...+jl =j

∂ l+j f

×

+ (εµ)2

∞ l=0

(x0 ))

∂xi1 . . . ∂xil ∂xj11 . . . ∂xjss

(εµ)l

x − x0 ) εµ

Kie1 ...irl ,

(i1 ...il ) (j1 ...js ), j1 +...+jl =j

×

∂ l+j f

∂xi1 . . . ∂xil ∂xj11 . . . ∂xjss

(x0 ).

j1 ...js (

x − x0 ) εµ

(5.7.15)

284


Here ˜1 (x0 ) ˜ ). ˜ir ...i ( x (5.7.16) ) + ∆N (N Nie1 ...irm (z) = N 1 l ε 0 ˜ ˜ where z = x−x εµ , ∆N (N ) is defined in terms of the functions N with multiindices (i1 . . . il of length less than l or with multi index ( (i1 . . . il but with superscripts not exceeding r − 1. An analogue of the representations (5.7.16) can also be found for the matrices Mi1 ...il , j1 ...js .

A formal asymptotic solution of the problem near a pre-nodal component Πx0 corresponding to the node x0 /ε is sought in the form ∞ ∞ ∂lV x − x0 (∞) ) (x0 ) u ˜ Πx ∼ ( εl µr (N Ni01 ...irl ( 0 εµ ∂xi1 . . . ∂xil r=0 l=0

(i1 ...il )

+ε2

Mi01 ...irl ,

(j1 ...js ), j1 +...+jl =j

j1 ...js (

x − x0 ) εµ

∂ l+j f

×

(x0 )) ∂xi1 . . . ∂xil ∂xj11 . . . ∂xjss x − x0 (∞) )ue ; + χe (x)ˆ ρ(αe εµ

(5.7.17)

e(x0 )

here the summation in the last sum in (5.7.17) extends over all segments e having x0 /ε as an end-point, χe (x) is the characteristic function of the section of e, αe the matrix of the transformation α which corresponds to e, and ρˆ(t) = ρ(t1 ) a differentiable function that vanishes for |t1 | ≤ c0 , is equal to unity for |t1 | ≥ c0 + 1, and such that 0 ≤ ρˆ(t1 ) ≤ 1, where t = (t1 , . . . , ts ). We set z = (x − x0 )/(εµ) and denote by Πx0 ,z,0 the image of Πx0 . Let ˜ α ˜ Bhj be a cylinder whose intersection with Πx0 is nonempty. We extend it to a semi-infinite cylinder whose basis has no common points with Πx0 . We denote α by Bhj∞ this extended cylinder, and by Πx0 ∞ the union of all such cylinders α Bhj∞ and Πx0 . Moreover, we denote by Πx0 ,z,∞ the image of Πx0 ,∞ under the transformation z = (x − x0 )/(εµ). Let χ ˜e (z) be the characteristic function of the dilated under transformation α z = (x − x0 )/(εµ) domain Bhj∞ corresponding to the section e. Substituting (5.7.17) into (5.7.1), (5.7.2), we find that Ni01 ...irl is the solution to the problem (similar to (4.3.7) of Chapter 4): Lz Ni01 ...irl =

e(x0 )

χ ¯e (z)(Lz ((1 − ρˆ(αe z))N Nie1 ...irl (z))

−(1 − ρˆ(αe z))Lz Nie1 ...irl (z)), z ∈ Πx0 ,z,∞ ,

5.7.

285


∂ ∂ ((1 − ρˆ(αe z))N Nie1 ...irl (z)) Ni01 ...irl = χ ¯e (z)( ∂νz ∂nz e(x0 )

−(1 − ρˆ(αe z))

∂ N e r (z)), z ∈ ∂Πx0 ,z,∞ . ∂nz i1 ...il

(5.7.18)

r . These problems as well as Similar problems are obtained for Mi01 ...i l , j1 ...js problem (5.7.18) are of the form Lz N 0 = χ ¯e (z)Lz φe (z) + φ0 (z), z ∈ Πx0 ,z,∞ . e(x0 )

s ∂ ∂ 0 φe (z) + ni φi e (z) , z ∈ ∂Πx0 ,z,∞ , (5.7.19) N = χ ¯e (z) ∂nz ∂nz i=1 e(x0 )

where φe are smooth functions and φ0 and φi e piecewise smooth functions with compact support. Lemma 5.7.1. Problem (5.7.19) is solvable in the class of functions that stabilize exponentially to a constant (in the sense of [88]-[90]) if and only if φ0 (z)dz = Πx0 ,z,∞

=

e(x0 )

s

nk φk e (z) ds +

∂ Πx0 ,z,0 \∂Πx0 ,z,∞

∂ Πx0 ,z,∞ k=1

∂ φe (z) ds ∂νz

where (n1 , ..., ns ) is the outward normal to Πx0 ,z,0 . The proof of the lemma is fully analogous to that of the main theorem in [89]. Lemma 5.7.1 and (5.7.17) yield the conditions ∂ (∞) ∆x uN dz = εµ χ ¯e (z)(1 − ρˆ(αe z)) e ∂νz Πx0 ,z,∞ e(x0 )

= −

e(x0 )

χ ¯e (z)

∂ Πx0 ,z,0 \∂Πx0 ,z,∞

M (∞)

and similar conditions for ue second sums in (5.7.16). Taking into account the fact

∂ N u ∂nz e

(∞)

N (∞)

; here ue

∆x ueN ∂ N u ∂nx e

(∞)

(∞)

∼

∼

ds =∼ 0, M (∞)

and ue

d2 v˜ , d˜ x21

d˜ v , d˜ x1

(5.7.20)

are the first and

286

CHAPTER 5. LATTICE STRUCTURES M (∞)

M (∞)

and and ∂n∂ x ue and considering the analogous expansions for ∆x ue e r e r ˜ ˜ Remark 2.2.3, we find that N and satisfy at the nodes the M i1 ...il i1 ...il , j1 ...js conditions (following from (5.7.20)) ˆN ˜ e r |x˜ (x )/ε = Ψr Λ i1 ...il 1 0 N, r ˆM ˜e r Λ ˜1 (x0 )/ε = ΨM, i1 ...il ,j1 ...js |x

i1 ...il (x0 /ε),

i1 ...il ,j1 ...js (x0 /ε),

(5.7.21)

˜ e q and M ˜e q where the right-hand sides of (5.7.21) are defined in terms of N i1 ...il i1 ...il , with indices (l1 , q) < (l, r) (that is, l1 ≤ l, q ≤ r and l1 = l or q = r), ê = Λ

Λe mesβe ,

j1 ...js

(5.7.22)

e(x0 )

Λe = ∂∂ξ˜ and ξ˜ = α(ξ − ξ0 ), ξ0 = x0 /ε. 1 If (5.7.21) holds then, by Lemma 5.7.1, problems (5.7.18) have solutions that stabilize exponentially, as |z| → ∞, to some ”limit” constants. For every α segment e(x0 ) and its corresponding half-bounded cylinder Bhj∞ , denote the limit constant Ce . From the representation (5.7.16) it follows that if we change the value of ˜ e q |x˜ (x )/ε by δ N ˜ e q |x˜ (x )/ε then the value of Ce , also changes by N i1 ...il 1 0 i1 ...il 1 0

˜ e q |x˜ (x )/ε δN i1 ...il 1 0 (as in section 4.3) ˜e q The value of M ˜1 (x0 )/ε changes according the same rule. i1 ...il , j1 ...js |x Suppose that for any e1 (x0 ) and e2 (x0 ), at the point x ˜1 (x0 )/ε we have ˜ e1 q |x˜ (x )/ε = N ˜ e2 q |x˜ (x )/ε , N i1 ...il 1 0 i1 ...il 1 0 ˜ e1 q M i1 ...il ,

˜1 (x0 )/ε j1 ...js |x

˜ e2 q =M i1 ...il ,

˜1 (x0 )/ε , j1 ...js |x

(5.7.23)

and that under these conditions, for every e = e(x0 ) the solution of (5.7.18) stabilizes to CeN , and the solution of the corresponding problem for Mi01 ...il , j1 ...js α stabilizes to CeM on the cylinder Bhj∞ corresponding to e. Then we redefine the e q e q ˜ ˜ values N | and M ˜1 (x0 )/ε ˜1 (x0 )/ε , so that their corresponding i1 ...il , j1 ...js |x i1 ...il x solutions of the problems in Πx0 ,z,∞ stabilize to one and the same constant for all segments e(x0 ) with the same end-point x0 . To this end , it suffices to set ˜ e q |x˜ (x )/ε and M ˜e q the new (improved) values of N ˜1 (x0 )/ε , equal i1 ...il 1 0 i1 ...il , j1 ...js |x N M to the old ones minus Ce and Ce , respectively. In other words, (5.7.23) needs to be replaced by conditions of the form ˜ e1 q |x˜ (x )/ε + δ N ˜ e1 q = N ˜ e2 q |x˜ (x )/ε + δ N ˜ e2 q , N i1 ...il 1 0 i1 ...il i1 ...il 1 0 i1 ...il ˜ e1 q M i1 ...il ,

˜ e1 q ˜1 (x0 )/ε +δ Mi1 ...il , j1 ...js j1 ...js |x

˜ e2 q =M i1 ...il ,

(5.7.24)

˜ e2 q ˜1 (x0 )/ε +δ Mi1 ...il , j1 ...js , j1 ...js |x

5.7.

287


for any segments e1 and e2 having the common end-point x0 . Here the constants ˜ e q and δ M ˜e q δN i1 ...il i1 ...il , j1 ...js are chosen so that (5.7.18) has solutions that stabilize to zero; in what follows we consider precisely these solutions of (5.7.18) M ˜e q ˜ e q = CeN and δ M (here δ N i1 ...il i1 ...il , j1 ...js = Ce ). From the above arguments we derive the following algorithm for the recursive 0 q 0 q ˜e q , M ˜e q computation of the matrices N i1 ...il i1 ...il , j1 ...js , Ni1 ...il and Mi1 ...il , j1 ...js . e 0 e 0 ˜ ˜ = 0; for q < 0, we The basis of the recursive process is N = 1, M ∅ ∅ e q e q 0 0 ˜ ˜ formally set Ni1 ...il = Mi1 ...il , j1 ...js = 0; Ni1 ...il are the solutions of problems (5.7.18) from which 0 Ni01 ...i = − l

0 Mi01 ...i l,

j1 ...js

= −

x ˜1 (x0 ) + const, ε

0 ˜ie ...i χ ¯e (z)ˆ ρ(αe z)N 1 l

0 ˜ ei ...i χ ¯e (z)ˆ ρ(αe z)M 1 l,

e(x0 )

e(x0 )

j1 ...js

x ˜1 (x0 ) + const. ε

0 q 0 q ˜e q , M ˜e q Afterwards, the functions N i1 ...il i1 ...il , j1 ...js , Ni1 ...il and Mi1 ...il , j1 ...js . are determined successively, in alphabetical order, by induction on (m + q, q) from (5.7.13),(5.7.14),(5.7.21),(5.7.24) and (5.7.18). Problems (5.7.13), (5.7.21) ˜ e q are written in the form: (5.7.24) for N i1 ...il

˜e q ) ˜e q ) d(N d2 (N i2 ,...,il i1 ,...,il e ˜e q + = −2γ − γie1 γie2 N i1 i3 ,...,il dξ˜2 dξ˜1 1

q +hqi1 ...il + Ψei1 ,...,i (ξ˜1 ) l

(5.7.25)

for any e, and

e(ξ0 )

mes βe

*

˜e q dN i1 ,...,il ˜e q + γie1 N i2 ,...,il dξ˜1

6

q = Ψξi10,...,i l

(5.7.26)

and ˜ e1 q = N ˜ e2 q + δ e1 ,e2 q N i1 ,...,il i1 ,...,il i1 ,...,il

(5.7.27)

in the nodes ξ0 for any e1 and e2 with the end-point in ξ0 . q q ,e2 q Here Ψei1 ,...,i (ξ˜1 ) are determined by Njq1 with q1 < q and Ψξi10,...,i and δie11,...,i l l l q1 e e are also determined by Nj with q1 < q, (γ1 , ..., γs ) are the directing cosines of the vector e with the initial point in ξ0 . Constants hqi1 ...il are chosen in such a way that every problem (5.4.25)-(5.4.27) has a 1−periodic solution. The possibility of such choice is a consequence of the following lemma. Lemma 5.7.2 Let Ψ(ξ˜1 ) be any 1-periodic function defined on B - which has generalized derivative along e for every e ⊂ B, - which is continuous on B and such that Ψe (ξ˜1 )dξ˜1 = 0, (5.7.28) e

e∩Q

288


where the summation extends over all segments e of the periodicity cell Q = (0, 1)s (here the parts of the segments e lying on the boundary of Q and differing by a period are taken only once). Then the following inequality holds: (Ψe (ξ˜1 ))2 dξ˜1 ≤ c (dΨe /dξ˜1 (ξ˜1 ))2 dξ˜1 (5.7.29) e

e∩Q

e∩Q

e

with a constant c depending only on B. Proof. Condition (5.7.28) and continuity of Ψ yield that there exist a point ξ 0 ∈ B ∩ Q, (remind that Q = (0, 1)s ) such that Ψ(ξ 0 ) = 0. Then applying the Newton-Leibnitz formula for any connected path Pξ0 ,ξ¯ of segments in B ∩ Q, ¯ with the initial point in ξ 0 and the final point in ξ, dΨ ¯ dl = Ψ(ξ) Pξ0 ,ξ¯ dl

we obtain the estimate

Pξ0 ,ξ¯

2

Ψ dl ≤ |P Pξ0 ,ξ¯|

Pξ0 ,ξ¯

dΨ dl

2

dl,

where |P Pξ0 ,ξ¯| is the length of the path Pξ0 ,ξ¯. Adding these inequalities for some pathes covering B ∩Q, we obtain estimate (5.7.29). Applying Lemma 5.7.2 and the Riesz theorem on representation of linear functionals in Hilbert space, we obtain the solvability of problems(5.7.25)(5.7.27) if and only if ˜e q ) d(N i2 ,...,il ˜ e q ) + hq − γie1 γie2 N − γie1 i3 ,...,il i1 ...il + ˜ d ξ e∩Q 1 e

e q ˜ +Ψ (ξ1 ) dξ˜1 mes βe + Ψξ0 q = 0. (5.7.30) i1 ,...,il

Here in integrals

i1 ,...,il

¯ ξ0 :ξ0 ∈Q

we choose any of the two ends of the delated segment e∩Q e as an end-point ξ0 to define ξ˜ as α(ξ − ξ0 )). In the sum ξ0 :ξ0 ∈Q¯ periodic points of the boundary of Q are taken only once. Because of equations (5.7.25), we obtain that (5.7.10) (and respectively (5.7.1)) is transformed into the homogenized equation ∞ ∞

εm−2 µq

m=2 q=0

+

∞ ∞

m=1 q=0

εm µq

q hN i1 ...im

(i1 ...im )

(i1 ...im ) (j1 ...js ):j1 +...+js =j

q hM i1 ...im ,

∂mV ∂xi1 . . . ∂xim

j1 ...js

∂j f ∂m ∼ j 1 ∂xi1 . . . ∂xim ∂x1 . . . ∂xjss

5.7.

CONDUCTIVITY OF A LATTICE: AN EXPANSION ∼ 0,

289 (5.7.31)

where the main part of the first sum is s

h0i1 i2

i1 ,i2 =1

where

h0i1 i2

∂2 V, ∂xi1 ∂xi2

are defined by relations

e⊂Q1 ∩B

mes

e

h0i1 i2 =

βe γie1

e⊂Q1 ∩B

˜ e 0) d(N i2 dξ˜1

+

γie2

mes βe |e|

dξ˜1

,

where Ni02 is a 1-periodic continuous solution to the cell problem ˜ e 0 + ξi ) d2 (N 2 i2 = 0; 2 ˜ dξ

(5.7.32)

1

for any e, and

mes βe

e(ξ0 )

and

˜ e1 N i2

˜ e 0 + ξi ) d(N 2 i2 =0 dξ˜1 0

˜ e2 =N i2

0

(5.7.33)

(5.7.34)

in the nodes ξ0 for any e1 and e2 with the end-point in ξ0 . Thus h0i1 i2 =

e⊂Q∩B

e

mes βe

e⊂Q∩B

˜ e 0 +ξi ) d(N i1 1 dξ˜1 dξ˜1

mes βe |e|

(5.7.35)

˜ e 0 are linear on every e. This formula coincides with the algorithm for and N i1 calculation the homogenized operator according to the principle of splitting in section 8.3 of [16]. Denote ˆ= L

s

i1 ,i2 =1

h0i1 i2

∂2 . ∂xi1 ∂xi2

ˆ has a symmetric positive definite Lemma 5.7.3 The homogenized operator L 0 0 matrix h = (hi1 i2 )i1 ,i2 =1,...,s . Proof. Applying the variational formulation to problem (5.7.32)-(5.7.34), we obtain

290


e⊂Q∩B

so

mes βe

e∩Q

˜e 0 ˜ e 0 + ξi ) dN d(N 1 i2 i1 dξ˜1 = 0, dξ˜1 dξ˜1

h0i1 i2 = =

e⊂Q∩B

e∩Q

mes βe

˜ e 0 +ξi ) ˜ e 0 +ξi ) d(N d(N i2 i1 2 1 dξ˜1 dξ˜1 dξ˜1

e⊂Q∩B

and so

mes βe |e|

,

(5.7.36)

h0i1 i2 = h0i2 i2 , i1 , i2 ∈ {1, ..., s}.

Moreover, for any η = (η1 , ..., ηs )T ∈ IRs ,

e⊂Q∩B

=

e⊂Q∩B

e∩Q

mes βe dξ˜1

e∩Q

s

h0i1 i2 ηi2 ηi1 =

i1 ,i2 =1

s

˜ e 0 + ξi )ηi 2 d i1 =1 (N 1 1 i1 mes βe dξ˜1 ≥ 0. dξ˜1

Let it be zero. Then for any e, s ˜ e 0 + ξi )ηi d i1 =1 (N 1 1 i1 = 0; dξ˜1

since Nie1 0 is a continuous function, for any e, s

i1 =1

˜ie 0 + ξi )ηi = const (N 1 1 1

and since B is a connected 1-periodic set, it is the same constant for all e. s Moreover, since i1 =1 Nie1 0 ηi1 is a linear 1-periodic function, it is a constant. s So, ξ η is a constant on B. It is equal to zero for ξ = 0, therefore s i1 =1 i1 i1 i1 =1 ξi1 ηi1 = 0. Taking now for any k ∈ {1, ..., s}, ξ = (δ1k , ..., δsk ), we prove that ηk = 0. Thus, η = 0, and h0 is a positive definite matrix. Lemma is proved. 0 q 0 q ˜e q , M ˜e q Thus, after all the functions N i1 ...il i1 ...il , j1 ...js , Ni1 ...il and Mi1 ...il , j1 ...js . are constructed, we can proceed with the construction of V, the formal asymptotic solution of the system of equations (5.4.12), which we seek in the form of a regular series in ε and µ : V =

∞

εp1 µp2 Vp1 p2 (x).

(5.7.36)

p1 ,p2 =0

where the coefficients of this series are solutions of the recursive sequence of problems ˆ Vp p = fp p (x). LV (5.7.37) 1 2 1 2

5.8.

HIGH ORDER HOMOGENIZATION OF ELASTIC LATTICES

291

where fp1 p2 (x) are defined in terms of the Vq1 q2 with indices (q1 , q2 ) < (p1 , p2 ) that is, qi ≤ pi , i = 1, 2 and q1 = p1 or q2 = p2 , and f00 = f (x); thus, the above equation coincides with the homogenized one obtained in section 5.1 by means of principle of splitting of the homogenized operator. It can be proved by induction that the fp1 p2 (x) are T −periodic in x and that fp1 p2 (x)dx = 0; (0,T )s

consequently, the following assertion holds Lemma 5.7.3. The sequence of problems (5.7.37) is recursively solvable in class of T −periodic functions Vq1 q2 with zero averages, that is, such that V (x)dx = 0. (0,T )s p1 p2 We denote by u(J) the truncated (partial) sum of the asymptotic series (5.7.5)-(5.7.7), (5.7.17), (5.7.36), obtained by neglecting the terms of order O(εJ + µJ ) (in the H 1 (BT )−norm, BT = Bµε ∩ (0, T )s ). Choosing J sufficiently large, we can show for any K that equation (5.7.1) and condition (5.7.2) are satisfied with remainders of order O(εK + ε−1 µK ) (in the L2 (BT )−norm). Hence , using the a priori estimate for problem (5.7.1),(5.7.2) given by Theorem 5.7.2, we obtain the following assertion. Theorem 5.7.3.If there are numbers α1 and α2 such that ε = O(µα1 ) and µ = O(εα2 ) as ε, µ → 0, then u − u ˜(J) H 1 (BT ) = O(εJ + µJ ), where u ˜(J) = u(J) − < u(J) >T , u is the solution of problem (5.7.1),(5.7.2), and < u(J) >T = 0, < . >T = .dx. BT

Thus,(5.7.5)-(5.7.7), (5.7.17), (5.7.36) is an asymptotic approximation of the solution of problem (5.7.1),(5.7.2). Remark 5.7.1 From the structure of the formal asymptotic solution (5.7.5)(5.7.7), (5.7.17), (5.7.36) it follows that, as ε, µ → 0, the solution converges in the norm .H 1 (BT ) / mesBT

to a function V00 independent of ε and µ. This function is a T −periodic solution to the homogenized problem ˆ V00 = f (x). LV

5.8

High order homogenization of elastic lattices

We consider the system of equations of elasticity theory in Bµ,ε

292


s ∂u ∂ = f (x), x ∈ Bµ,ε , Aij Lx u = ∂xj ∂xi i,j=1

(5.8.1)

∂u = 0, x ∈ ∂Bµ,ε , ∂ν

(5.8.2)

where

s ∂u ∂u ni , = Aij ∂x ∂ν j i,j=1

and Aij are constant s × s matrices, the elements akl ij of the Aij satisfy the relations akl ij = (δij δkl + δil δjk )M + λδik δjl , Here M and λ are the Lame´ coefficients, (n1 , . . . , ns ) is the normal to ∂Bµ,ε , f ∈ C ∞ is a T −periodic s−dimensional vector-valued function, T is a multiple of ε. As above s = 2 or s = 3. We seek the solution u(x) in the class of T −periodic s−dimensional vector-valued functions of HT1 −per (Bµ,ε ). Here HT1 −per (Bµ,ε ) stands for the completion of the set of T −periodic differentiable functions defined at Bµ,ε with respect to the H 1 (Bµ,ε ∩ (0, T )s )−norm. Below we construct the asymptotic expansion of the solution to problem (5.8.1)-(5.8.2) as µ, ε → 0 under some assumptions P F1 and P F2 of geometrical rigidity of the lattice structure. These assumptions will be formulated below for the limit problem. We assume as well that the following integrals of the right hand-side vanish: f (x)dx = 0, f (x)dx = 0. (0,T )s

Bµ,ε ∩(0,T )s

Theorem 5.8.1. There exist a unique solution to problem (5.8.1),(5.8.2) such that u(x)dx = 0. Bµ,ε ∩(0,T )s

and the estimate holds: there exist constants C and q independent of µ and ε such that −q u2H 1 f 2L2 (Bµ,ε ∩(0,T )s ) . (Bµ,ε ∩(0,T )s ) ≤ Cµ T −per

This means that any solution of problem (5.8.1),(5.8.2) satisfies the estimate ∇u2L2 (Bµ,ε ∩(0,T )s ) ≤ Cµ−q f 2L2 (Bµ,ε ∩(0,T )s ) . The result on existence and uniqueness of solution to a mixed boundary value problem of elasticity is well known (see for example [55]). Let us prove the a priori estimate. As it was shown in section 5.4, the Korn inequality constant for structures depends on the small parameter. Composing the example of Lemma 5.4.1 for every rod, we obtain an ε−periodic function

5.8.


293

that has the Korn inequality constant of order µ−1 . On the other hand the theorem of Appendix shows that this Korn inequality constant is not worse é inequality constant is also not worse than µ−q (q ∈ IR). Moreover, the Poincar´ than some power of µ. So, finally we obtain the estimate of Theorem 5.4.1. The asymptotic of the solution to problem (5.8.1), (5.8.2) are constructed in several steps. These steps are similar to the procedure of the previous section. The new elements are: the presence in the asymptotic expansion of the term of order ( µε )2 (see section 5.6) and the dependency of the Korn inequality constant on µ−q (this dicrepancy is not dangerous for the justification of the truncated asymptotic expansion because we apply the same improving of the accuracy procedure as in section 4.4).

1. First , we construct the asymptotic corresponding to a segment e = enh in B . We change to new variables x ˜ = α(x − εh). In these new variables, the ˜ α } becomes the rod. cylinder {x/ε ∈ B hj ˜ < lε, x ˜′ /εµ ∈ βj }; Uεµ = {˜ x1 ∈ IRs | 0 < x we set Γεµ = {˜ ˜ < lε, x ˜′ /εµ ∈ ∂β x1 ∈ IRs | 0 < x βj }. In what follows β = βj . We write f˜(˜ x) = αf (α∗ x ˜ + εh) and u ˜(˜ x) = ∗ αu(α x ˜ + εh) . ˜ α becomes Then problem (5.8.1), (5.8.2) on B hj Pu ˜=−

s ∂u ˜ ∂ x), x ) = f˜(˜ ˜ ∈ Uεµ , (Aij ∂x ∂x j i i,j=1

s ∂u ˜ ∂u ˜ ni = 0, x ˜ ∈ Γεµ . Aij = ∂xj ∂ν i,j=1

(5.8.3)

(5.8.4)

We make the substitution ξ ′ = (ξ˜2 , . . . , ξ˜s ), ξ˜i = x ˜i /(εµ) and expand f˜ in an asymptotic series with respect to εµξ˜i , of the form f˜ ∼

∞ j=0

(εµ)

j

rj

ΦF Fjr (ξ˜2 , . . . , ξ˜s )ψ˜jr (˜ x1 ),

r=1

In this formula the d-dimensional vectors ψ˜jr (˜ x1 ) are the values of the qs−1 ∂xs j−q1 −...qs−1 at the point x = α∗ (˜ x, 0, . . . , 0) + derivatives ∂ j f /∂xq11 ...∂xs−1 εh, Fjr (ξ ′ ) are matrices of polynomials and Φ is a matrix of rigid displacements. In particular , ψ˜00 (˜ x1 ) = (f˜(˜ x1 , 0, 0), 0)∗ for s = 3, and ψ˜00 (˜ x1 , 0) = f˜(˜ x1 , 0) for s = 2, , F00 = I (the identity matrix ) and r0 = 1. We represent Fjr in the form F¯jr + F˜jr , where Φ∗ ΦF˜jr β = 0 and F¯jr = Φ∗ ΦF Fjr β .

294


Using superposition , we seek the solution to problem (5.8.3) , (5.8.4) in the form: u ˜∼

(∞)

(εµ)j

rj

u ˜jr ,

(5.8.5)

r=1

j=0

where u ˜jr is the solution of the problem for a right hand side of the form ˜ ΦF Fjr ψjr . From now on in this section we omit the ∼ that denotes the change of variables, fix j and r and do not mention them explicitly when this does not create ambiguity . We seek a formal asymptotic solution to problem (5.8.3) , (5.8.4) with right hand side ΦF Fjr ψjr in the form of a series

u ˜∞ jr ∼

∞

(εµ)l Nl

l=0

x ˜( ˜ ) ˜ ′ dl ω 1

εµ

dxl1

+ (εµ)2

∞

(εµ)l Ml

l=0

x x1 ) ˜′ dl ψ˜jr (˜ l εµ d˜ x1

(5.8.8)

where Nl and Ml are the matrix valued functions of section 2.2; ω ˜ (˜ x1 ) = v˜εµ (x)+ ε2 w ˜εµ (x), vεµ (x) = α∗ v˜εµ (x), and wεµ (x) = α∗ w ˜εµ (x) are defined on IRs , and

v˜εµ (x) ∼ E0 αV (x)+

w ˜εµ (x) ∼

∞ ∞

εm µq

m=1 q=0

∞ ∞

m=0 q=0

(i1 ...im ): in ∈{1...s}

εm µq

∂ m V (x) x˜1 (x) ˜q ) N i1 ...im ( ε ∂xi1 . . . ∂xim

(j1 ...js ): j1 +...+js =j (i1 ...im ): in ∈{1...s}

∂j f ∂m x˜1 (x) ˜q ) ×M i1 ...im ,j1 ...js ( ∂xi1 . . . ∂xim ∂xj11 . . . ∂xjss ε

(5.8.9)

˜ ˜ ˜q ˜q Here V (x) is an s-dimensional vector N i1 ...im (ξ1 ) and Mi1 ...im ,j1 ...js (ξ1 ) are d × s matrices of the form ˜q ˜ q(4) + µ−2 N ˜ q(2) ˜ q(1) + µ−1 N N ˆ Niq1 ...im α∗ = N i1 ...im = αN i1 ...im i1 ...im i1 ...im q ∗ ˜q M i1 ...im ,j1 ...js = αMi1 ...im ,j1 ...js α

−1 ˜ q(4) ˜ q(2) ˜ q(1) Mi1 ...im ,j1 ...js + µ−2 M =M i1 ...im ,j1 ...js + µ i1 ...im ,j1 ...js ,

(5.8.10)

5.8.


E0 = I for s = 2, E0 =

0

I 0

295

0

for s = 3 (here I is the unit s × s matrix); α ˆ = α for s = 2, and ⎛ ⎞ 0 ⎜ α 0⎟ ⎟ α ˆ=⎜ ⎝ 0⎠ 0 0 0 1

for s = 3, and x ˜1 (x)/ε is the projection of x/ε on e ; all the rows of the matrices ˜ q(1) and M ˜ q(1) ˜ q(4) N i1 ...im i1 ...im ,j1 ...js except the first one, all the rows of Ni1 ...im and ˜ q(4) ˜ q(2) and M except the fourth one for s = 3, and all the rows of N i1 ...im ,j1 ...js

i1 ...im

˜ q(2) M i1 ...im ,j1 ...js except the second one ( and the third one for s = 3) , are zero. We remark that ψ˜jr can be represented in the form

ψ˜jr ( ˜1 (x)) =

(j1 ...js ): j1 +...+js =j

∂ j f (x) ˜ j ...j ( x ) , K 1 s j ε ∂x1 1 . . . ∂xs js

(5.8.11)

˜ j ...j are constant for every e (this constant changes if one passes from where K 1 s one segment e to another). The substitution of (5.8.8) in (5.8.3) and (5.8.4) is described in section 2.2 (with µ replaced by εµ ). As a result, we find that (5.8.4) is satisfied asymptotically exactly and that (5.5.3) yields a system of d equations ”homogenized over the rod” , of the form hN 2

+

∞

d4 ω ˜ d2 ω ˜ + (εµ)2 hN − F jr ψ˜jr 4 2 d˜ x1 d˜ x41

(εµ)l−2 hN l

l=6

∞

dl ω ˜ dl ψ˜jr ∼ 0. + (εµ)l hM l l d˜ xl1 d˜ x1 l=0

(5.8.12)

Here we have not yet used the fact that ω has the form (5.8.9). We substitute (5.8.9) in (5.8.12) multiplied on the left by α ˆ∗ and gather together the coefficients of the same powers of εµ and the derivatives Di V =

∂j f ∂mV . , Dj f = j ∂xi1 . . . ∂xim ∂x11 . . . ∂xjss

We obtain an asymptotic equality of the form ∞ ∞ x ∂mV ˜1 (x) q m−2 q ε µ HiN1 ...i m ∂xi1 . . . ∂xim ε m=1 q=0 (i1 ...im )

296


+

(j1 ...js ):j1 +...+js =j

HiM1 ...iq m , j1 ...js

x ∂m ∂j f ˜1 (x) j ε ∂xi1 . . . ∂xim ∂x11 . . . ∂xjss

∼0

(5.8.13)

q ˜ ˜q HiN1 ...i =α ˆ e∗ RN ˆ e∗ Tiq1 ...im (N )αe , i1 ...im (ξ1 )αe + α m

HiM1 ...iq m ,

j1 ...js

˜q =α ˆ e∗ RM i1 ...im ,

˜

j1 ...js (ξ1 )αe

+α ˆ e∗ Tiq1 ...im ,

j1 ...js (M)αe ,

(5.8.14)

Tiq ...i (N )(T Tiq1 ...im , j1 ...js (M)) are linear combinations of the Niq1 ...im ˜q 1 m (M i1 ...im , j1 ...js ) and their derivatives with multi-index (i1 . . . im ) of length m1 less than m or with the multi-index (i1 , ..., im ) with m = m1 and with superr˜1 (x)/ε, s cript q1 < q, and R is a differential operator along the variable ξ˜1 = x of the form ⎞ ⎛ ¯ ∂ 22 E 0 ˜ ∂ ξ1 ⎠ R = ⎝ ¯ ξ˜2 β ∂ 44 0 −E 2 ˜ ∂ξ 1

for s = 2,

⎛

⎜ ⎜ ⎜ R = ⎜ ⎜ ⎝

¯ ∂ 22 E ∂ ξ˜ 1

0

0 4 ˜ ¯ −Eξ22 β ∂∂ξ˜4 1

0

0

0

0

0 4 ˜ ¯ −Eξ32 β ∂∂ξ˜4

0

1

0

0

0

⎞

0 2 ¯ M ∂∂ξ˜2 1

⎟ ⎟ ⎟ ⎟ ⎟ ⎠

for s = 3. ˜ N (ξ1 ) and H ˜M We equate H i1 ...is i1 ...is , j1 ...jm (ξ1 ) with the constant matrices q q hi1 ...im and hi1 ...im , j1 ...js so that for s = 3 the fourth row and the fourth column are zero. We denote by hqi1 ...im , and hqi1 ...im , j1 ...js the minors of these matrices obtained by removing their fourth rows and columns. We obtain a homogenized equation for V of the form: ∞ ∞

m=1 q=0

+

∞ ∞

m=1 q=0

εm−2 µq

εm−2 µq

hqi1 ...im

(i1...im)

∂mV ∂xi1 . . . ∂xim

hqi1 ...im ,

(i1 ...im ) (j1 ...js ): j1 +...+js =j

∼ 0.

j1 ...js

∂j f ∂m ∂xi1 . . . ∂xim ∂xj11 . . . ∂xjss

(5.8.15)

5.8.


297

The coefficients hqi1 ...im , and hqi1 ...im , j1 ...js of this equation will be determined later. ˜q ˜q Thus the N i1 ...im and Mi1 ...im , j1 ...js satisfy the equations

˜ q i ...i = T˜ q ˆ ε hqi1 ...im α ˆ ε∗ RN 1 m i1 ...im (N ) + α ˜ q i ...i , RM 1 m (∞)

= T˜iq1 ...im ,

js ...js

js ...js (M )

(5.8.16)

+α ˆ ε hqi1 ...im α ˆ ε∗ .

(5.8.17)

(∞)

Let u ˜ε = u ˜jr (x) be the asymptotic series (5.8.8), (5.8.9), which is the ˜ α }. solution of (5.8.1) (5.8.2) in the cylinder {x : xε ∈ B hj We expand the functions

∂ l ψjr ∂j f ∂m ∂mV , , , j j 1 s ∂xi1 . . . ∂xim ∂xi1 . . . ∂xim ∂x1 . . . ∂xs ∂x ˜li

˜ i ...i ,j ...j ( x˜1 (x) ) in Taylor series in the neighborhood of ˜i ...i ( x˜1 (x) ) and M N s 1 m 1 1 m ε ε the point x0 corresponding to the node xε0 , and substitute these expansions in (5.8.8 ) . We obtain ∞ ∞ ∂lV x − x0 (x0 ) ) u ˜(∞) ∼ α( εl µr (N Nie1 ...irl ( ε ∂xi1 . . . ∂xil εµ r=−2 l=0

(i1 ...il )

+ε2

Mie1 ...irl ,

j1 ...js (

(j1 ...js ), j1 +...+jl =j

×

+ (εµ)2

∞

∂ l+jf

∂xi1 . . . ∂xil ∂xj11 . . . ∂xjss

(εµ)l

l=0

x − x0 ) εµ

(x0 ))

Kie1 ...irl ,

j1 ...js (

(i1 ...il ) (j1 ...js ), j1 +...+jl =j

×

∂ l+jf

∂xi1 . . . ∂xil ∂xj11 . . . ∂xjss

x − x0 ) εµ

(x0 )).

(5.8.18)

Let ˜1 = x ˜1 /(εµ). Then

˜1 (x0 ) ˜1 (x0 ) ˜ r+1 (4) ( x ˜ r(1) ( x ) )+N Nie1 ...irm (z) = α∗ N0 (˜′ ) N i1 ...il i1 ...il ε ε

˜ r+1 (2) x ∂N ˜1 (x0 ) x ˜1 (x0 ) i1 ...il )˜1 α ( )+ ε ε ∂ ξ˜1

˜ r+1 (2) x ∂N ˜1 (x0 ) i1 ...il ′ ˜ ). )α + ∆N (N ( +N N1 (˜ ) ε ∂ ξ˜1 r+2 (2)

˜ +N i1 ...il

(

(5.8.19)

298


˜ ) is defined in terms of the functions N ˜ with multi-indices (i1 . . . il ) where ∆N (N of length less than l or with multi index (i1 . . . il ) but with superscripts not r(2) r(4) exceeding r − 1, or in terms of Ni1 ...il , Ni1 ...il . An analogue of the representations (5.8.19) can also be found for the matrices r Mie1 ...i . l , j1 ...js A formal asymptotic solution of the problem near a pre-nodal component Πx0 corresponding to the node x0 /ε is sought in the form ∞ ∞ ∂lV x − x0 (∞) (x0 ) ) u ˜ Πx ∼ ( εl µr (N Ni01 ...irl ( 0 ∂xi1 . . . ∂xil εµ r=−2 l=0

(i1 ...il )

+ε2

Mi01 ...irl ,

(j1 ...js ), j1 +...+jl =j

j1 ...js (

x − x0 ) εµ

∂ l+jf

×

(x0 )) ∂xi1 . . . ∂xil ∂xj11 . . . ∂xjss x − x0 (∞) )ue ; + χe (x)ˆ ρ(αe εµ

(5.8.20)

e(x0 )

(∞)

(∞)

here ue = α∗ u ˜e , the summation in the last sum in (5.8.20) extends over all segments e having x0 /ε as an end-point, χe (x) is the characteristic function of the section of e, αe the matrix of the transformation α which corresponds to e, and ρˆ(t) = ρ(t1 ) a differentiable function that vanishes for |t1 | ≤ c0 , is equal to unity for |t1 | ≥ c0 + 1, and such that 0 ≤ ρˆ(t1 ) ≤ 1, where t = (t1 , . . . , ts ). As in the previous section we set z = (x−x0 )/(εµ) and denote by Πx0 ,z,0 the ˜ ˜ α be a cylinder whose intersection with Πx is nonempty. image of Πx0 . Let B 0 hj We extend it to a semi-infinite cylinder whose basis has no common points with α Πx0 . We denote by Bhj∞ this extended cylinder, and by Πx0 ∞ the union of α all such cylinders Bhj∞ and Πx0 . Moreover, we denote by Πx0 ,z,∞ the image of Πx0 ,∞ under the transformation z = (x − x0 )/(εµ). α Let χ ˜e (z) be the characteristic function of Bhj∞ corresponding to the section e. Substituting (5.8.40) in (5.8.1), (5.8.2), we find that Ni01 ...irl is the solution to the problem (similar to (4.3.7) of Chapter 4). Lz Ni01 ...irl = χ ¯e (z)(Lz ((1 − ρˆ(αe z))N Nie1 ...irl (z)) e(x0 )

−(1 − ρˆ(αe z))Lz Nie1 ...irl (z)), z ∈ Πx0 ,z,∞ , ∂ ∂ ((1 − ρˆ(αe z))N Nie1 ...irl (z)) Ni01 ...irl = χ ¯e (z)( ∂νz ∂νz e(x0 )

−(1 − ρˆ(αe z))

∂ N e r (z)), z ∈ ∂Πx0 ,z,∞ . ∂νz i1 ...il

(5.8.21)

5.8.


299

r Similar problems are obtained for Mi01 ...i . These problems as well as l , j1 ...js problem (4.8.21) are of the form Lz N 0 = χ ¯e (z)Lz φe (z) + φ0 (z), z ∈ Πx0 ,z,∞ . e(x0 )

s ∂ ∂ 0 φe (z) + ni φi e (z) , z ∈ ∂Πx0 ,z,∞ , (5.8.22) N = χ ¯e (z) ∂νz ∂νz i=1 e(x0 )

where φe are smooth functions and φ0 and φi e piecewise smooth functions with compact support. Let J˜(z) be the matrix in section 2.2 and let J be the linear span of columns of J˜(z). Lemma 5.8.1 Problem (5.8.22) is solvable in the class of vector-valued functions that stabilize exponentially to a vector-valued function in J (in the sense of [130]) if and only if J˜∗ (z)φ0 (z)dz =

e(x0 )

Πx0 ,z,∞

∂ Πx0 ,z,∞

J˜∗ (z)

s

k=1

∂ φe (z) ds nk φk e (z) ds + J˜∗ (z) ∂ν z ∂ Πx0 ,z,0 \∂Πx0 ,z,∞

where (n1 , ..., ns ) is the outward normal to Πx0 ,z,0 . The proof of the lemma is fully analogous to that of the main theorem in [130]. Lemma 5.8.1 and (5.8.20) yield the conditions ∂ ˜∗ (∞) χ εµ ¯e (z)(1 − ρˆ(αe z)) dz J (z)Lx uN e ∂ν z Πx0 ,z,∞ e(x0 )

= −

e(x0 )

∂ N ue χ ¯e (z)J˜∗ (z) ∂ν z ∂ Πx0 ,z,0 \∂Πx0 ,z,∞ M (∞)

N (∞)

; here α∗ ue and similar conditions for ue and second sums in (5.8.8), and

s ∂ ∂ ∂ = , Lx = Aij ∂ν ∂x ∂x z j i i,j=1

Taking into account the fact Lx ueN

∂ N u ∂νx e

(∞)

∼

αe∗

(∞)

∞ l=1

∼

(εµ)

∞

l−1

ds , M (∞)

and α∗ ue s

ni

i,j=1

(εµ)l−2 hN l

l=2

(∞)

∂ Aij ∂xj

(5.8.23) are the first

.

dl v˜ , d˜ xl1

dl v˜ ∂N Nl , + A11 Nl−1 A1j ∂ξξj d˜ xl1 j=1

s

300

CHAPTER 5. LATTICE STRUCTURES M (∞)

M (∞)

and and ∂ν∂x ue and considering the analogous expansions for Lx ue e r e r ˜ ˜ relation of Remark 2.3.2, we find that Ni1 ...il and Mi1 ...il , j1 ...js satisfy at the nodes the conditions (following from (5.8.23)) r ˆN ˜ie ...i Λ | ˜1 (x0 )/ε = ΨrN, 1 l x

ˆM ˜e r Λ ˜1 (x0 )/ε i1 ...il ,j1 ...js |x

=

i1 ...il (x0 /ε),

ΨrM, i1 ...il ,j1 ...js (x0 /ε),

(5.8.24)

˜ e q and M ˜e q where the right-hand sides of (5.8.24) are defined in terms of N i1 ...il i1 ...il , j1 ...js with indices (l1 , q) < (l, r) (that is, l1 ≤ l, q ≤ r and l1 = l or q = r), ê = Λ Γe Λe mesβe , (5.8.25) e(x0 )

Λe is the operator in (4.4.17); Γe the d × d matrix of the transformation J˜∗ (z)αe∗ = Γe J˜∗ (αe z), where J˜(z) is the matrix of rigid displacements of section 4.4, and βe the crosssection of the cylinder corresponding to e. As it was proved in Chapter 4 ⎞ ⎛ 0 αe∗ 0⎠ Γe = ⎝ 0 0 1 for s = 2, and

Γe =

αe∗ O

O ¯e Γ

(5.8.26)

¯ e is an orthogonal 3 × 3 matrix and O the 3 × 3 zero matrix. for s = 3, where Γ If (5.8.24) holds then, by Lemma 5.8.1, problems (5.8.21) have solutions that stabilize exponentially, as |z| → ∞, to some ”limit” functions in J . For α every segment e(x0 ) and its corresponding half-bounded cylinder Bhj∞ , the limit ˜ function can be represented in the form J (z)Ce , where Ce is a constant d¯ × s matrix with d¯ = 3 for s = 2 and d¯ = 6 for s = 3; here J˜(z) is a representative e q of a linear span J . On the other hand, we consider the d¯× s matrices SÑ , i1 ...il e q ˜ e r . and M ˜e r and S˜M, constructed from the matrices N i1 ...il i1 ...il , j1 ...js i1 ...il , j1 ...js e q as follows: for s = 2, the first two rows of the matrix SÑ , i1 ...il coincide with ˜ e q (1) + N ˜ e q+2 (2) , and the third row with the second row of the matrix N i1 ...il i1 ...il ∂ ˜ e q+1 N ; for s = 3, the first four rows of S˜e q coincide with the matrix N , i1 ...il

∂ ξ˜1 i1 ...il ˜ e q (1) + N i1 ...il

˜ e q+2 (2) + N ˜ e q+1 (4) ae , and the last two rows with the second and N i1 ...il i1 ...il ∂ ˜ e q+1 third rows of ∂ ξ˜ Ni1 ...il . Here ae is normalization factor (2.3.3). The matrices 1 e q ˜e r S˜M, i1 ...il , j1 ...js are constructed analogously from Mi1 ...il , j1 ...js . If we change e q e q the value of SÑ , i1 ...il |x˜1 (x0 )/ε by δ SÑ , i1 ...il , then the value of J˜(z)Ce , changes e q ˜e q by αe∗ J˜(αe z)δ SÑ , i1 ...il αe , that is, the value of Ce changes by Γe δ SN , i1 ...il αe . e q |x˜ (x )/ε changes according the same rule. The value of S˜ M, i1 ...il , j1 ...js

1

0

5.8.


301

Suppose that for any e1 (x0 ) and e2 (x0 ), at the point x ˜1 (x0 )/ε we have e1 q ˜ e2 q Γe1 SÑ , i1 ...il αe1 = Γe2 SN , i1 ...il αe2 , e1 q Γe1 S˜M, i1 ...il

j1 ...js αe1

e2 q = Γe2 S˜M, i1 ...il

j1 ...js αe2 ,

(5.8.27)

and that under these conditions, for every e = e(x0 ) the solution of (5.8.21) stabilizes to J˜(z)CeN , and the solution of the corresponding problem for Mi01 ...il , j1 ...js α stabilizes to J˜(z)CeM on the cylinder Bhj∞ corresponding to e. Then we redefine e q e q ˜ ˜ SN , i1 ...il |x˜1 (x0 )/ε and SM, i1 ...il , j1 ...js |x˜1 (x0 )/ε so that their corresponding solutions of the problems in Πx0 ,z,∞ stabilize to one and the same rigid displacement in J for all segments e(x0 ) with the same end-point x0 . To this end , it suffices to e q ˜e q set the new (improved) values of SÑ ˜1 (x0 )/ε and SM, i1 ...il , j1 ...js |x ˜1 (x0 )/ε , i1 ...il |x equal to the old ones minus Γ∗e CeN αe∗ and Γ∗e CeM αe∗ , respectively. In other words, (5.8.27) needs to be replaced by conditions of the form

Γe1

e1 q e1 q e2 q e2 q ˜ ˜ ˜ ˜ SN , i1 ...il + δ SN , i1 ...il αe1 = Γe2 SN , i1 ...il + δ SN , i1 ...il αe2 , e1 q S˜M, i1 ...il j1 ...js

Γe1

= Γe2

e2 q S˜M, i1 ...il

j1 ...js

+

e1 q δ S˜M, i1 ...il j1 ...js

e2 q + δ S˜M, i1 ...il

j1 ...js

αe1

αe2 ,

(5.8.28)

for any segments e1 and e2 having the common end-point x0 . Here the constant ¯ matrices δ S˜e q ˜ e2 q d×s N , i1 ...il and δ SM, i1 ...il j1 ...js are chosen so that (5.8.21) has solutions that stabilize to zero; in what follows we consider precisely these solutions e q ∗ N ∗ ∗ M ∗ ˜ e2 q of (5.8.21) ( here δ SÑ , i1 ...il = Γe Ce αe , and δ SM, i1 ...il j1 ...js = Γe Ce αe , e q e q and (5.8.28) contains now the ”new” values of SÑ , i1 ...il and S˜M, i1 ...il , j1 ...js ). From the above arguments we derive the following algorithm for the recursive 0 q 0 q ˜e q , M ˜e q computation of the matrices N i1 ...il i1 ...il , j1 ...js , Ni1 ...il and Mi1 ...il , j1 ...js . ˜ e 0 = I, M ˜ e 0 = 0; for q < 0, we The basis of the recursive process is N ∅ ∅ e q e q 0 −2 ˜ ˜ formally set Ni1 ...il = Mi1 ...il , j1 ...js = 0; Ni1 ...il are the solutions of problems (5.8.18), (5.8.19),(5.8.21) from which −2 Ni01 ...i = − l

−2 Mi01 ...i l,

j1 ...js

e(x0 )

= −

e 0 (2)

˜ χ ¯e (z)ˆ ρ(αe z)αe∗ N0 (αe z)N i1 ...il

e(x0 )

x ˜1 (x0 ) αe + const, ε

x ˜1 (x0 ) ˜ e 0 (2) αe + const. χ ¯e (z)ˆ ρ(αe z)αe∗ N0 (αe z)M i1 ...il , j1 ...js ε

0 q ˜e q , M ˜e q Afterwards, the matrix-valued functions N i1 ...il i1 ...il , j1 ...js , Ni1 ...il and are determined successively, in alphabetical order, by induction on

q . Mi01 ...i l , j1 ...js

302


(m+q, q) from (5.8.16),(5.8.17),(5.8.24),(5.8.28). Problems (5.8.16),(5.8.24),(5.8.28) ˜ e q split into recursively solvable pairs of problems: for N ˜ e q (1) αe , confor N i1 ...il i1 ...il sisting of the first equation (5.8.16), the s conditions

x0 ∗ e r r ˜ αe (5.8.29) αe Λe N ˜1 (x0 )/ε = ΨN , i1 ...il i1 ...il αe mesβe |x ε e(x0 )

¯ (we denote by Athe first s rows of the matrix A ) and

e1 r (1) e2 r (1) ∗ (2) ∗ (2) ˜ ˜ αe1 Ni1 ...il + λe1 αe1 |x˜1 (x0 )/ε = αe2 Ni1 ...il + λe2 αe2 |x˜1 (x0 )/ε , (5.8.30) (2) where λe is a linear d × s matrix-valued function on e whose first (and fourth, ˜ e q (2) + N ˜ e q (4) (for s = 2 the last term is for s = 3) row is zero, and for N i1 ...il i1 ...il missing), consisting of the remaining equations (5.8.16) and conditions (5.8.24), as well as conditions (5.8.28). A similar split also occurs for the problem for ˜e q M i1 ...il , j1 ...js αe . é- Friedrichs Suppose that the following Condition P F1 , analogous to the Poincar´ inequality, holds. ˜ ξ˜1 ) be any 1-periodic s-dimensional vector-valued Condition P F1 . Let Ψ( function defined on B - which has generalized derivative along e for every e ⊂ B, - which is such that all its components, except the first, are linear on every segment e, ˜ e1 = αe∗ Ψ ˜ e2 - which satisfies at all nodes x0 the matching conditions αe∗1 Ψ 2 for any two segments e1 (x0 ) and e2 (x0 ), having the common end-point x0 , and - which satisfies the condition ˜ e (ξ˜1 )dξ˜1 = 0, αe∗ Ψ (5.8.31) e

e∩Q

where the summation extends over all segments e of the periodicity cell Q = (0, 1)s (here the parts of the segments e lying on the boundary of Q and differing by a period are taken only once). Then it is assumed that the following inequality holds: ˜ e1 (ξ˜1 ))2 dξ˜1 ≤ c ˜ e1 /dξ˜1 (ξ˜1 ))2 dξ˜1 (Ψ (dΨ e

e∩Q

e

e∩Q

˜ e is the first component of the with a constant c depending only on B. Here Ψ 1 e ˜ vector-valued function Ψ . Thus, Condition P F1 is satisfied if for example only one node lies on the periodicity cell B ∩ Q. Lemma 5.8.2 Problem (5.8.16) (the first equation), (5.8.29), (5.8.30) is solvable if and only if

5.8.


e

e∩Q

303

mesβe (−αe∗ T˜iq1 ...il (N )αe + hqi1 ...il (N ))dξ˜1

+

x0 ∩Q

x0 ΨqN , i1 ...il ( )αe ε

= 0.

(5.8.32)

Here the summation in the second sum extends over all the nodes of the periodicity cell. The nodes lying on the boundary of Q and differing by a period are taken only once. The solution is determined up to an arbitrary additive constant s × s matrix. Proof follows from the Lax-Milgram theorem. Suppose that for s = 3 the following Condition P F2 , analogous to the Poincar´ é- Friedrichs inequality, holds. ˜ ξ˜1 ) be any three-dimensional 1-periodic vectorCondition P F2 . Let Ψ( valued function defined on B ˜ 1 has generalized derivative along e for every - whose first component Ψ segment e ⊂ B, ˜ k , k = 2, 3, have two generalized - whose second and third components Ψ derivatives along e for every segment e ⊂ B, - whose second and third components vanish at all nodes, and - which satisfies at all nodes x0 the matching conditions of the form ⎛

⎛ ˜ e1 ⎞ ˜ e2 ⎞ Ψ Ψ 1 1 ¯ ∗e ⎝ dΨ ¯ ∗e ⎝ dΨ ˜ e1 /dξ˜1 ⎠ = Γ ˜ e2 /dξ˜1 ⎠ Γ 2 2 2 1 ˜ e1 /dξ˜1 ˜ e2 /dξ˜1 dΨ dΨ 3 3

(5.8.33)

for any two segments e1 (x0 ) and e2 (x0 ), having the common end-point x0 . Then it is assumed that the following inequality holds:

˜e ˜e d Ψ d Ψ 3 2 2 2 e 2 ˜ 1 (ξ˜1 )) + ( (ξ˜1 )) dξ˜1 (ξ˜1 )) + ( Ψ dξ˜1 dξ˜1 e∩Q e

≤ c

e

e∩Q

˜e ˜e ˜e d2 Ψ d2 Ψ dΨ 3 ˜ 2 ˜ 1 ˜ 2 2 2 dξ˜1 (ξ1 )) + ( 2 (ξ1 )) + ( 2 (ξ1 )) dξ˜1 dξ˜1 dξ˜1

with a constant c depending only on B. We suppose also for simplicity of presentation that for any node there are three non-coplanar (if s = 3), respectively two non-collinear (if s = 2), segments e having this node as an end-point. This condition will be called the noncoplanarity condition. Lemma 5.8.3 If s = 3 and Condition P F2 holds, then the problem for ˜ e q (2) + N ˜ e q (4) (consisting of the last three equations (5.8.16), the last N i1 ...il i1 ...il three matching conditions (5.8.24), and matching conditions (5.8.28)) is also solvable.

304


Analogues of Lemmas 5.8.2 and 5.8.3 also hold for problems (5.8.17),(5.8.24), ˜e q (5.8.28) for M i1 ...il , j1 ...js . The solvability of the problems in Lemmas 5.8.2 and 5.8.3 is proved by means of the Riesz representation theorem for a bounded linear functional on a Hilbert space. An analysis of the right-hand sides of problems (5.8.17),(5.8.24),(5.8.28) ˜ e 0 (2) (N ˜ e 0 (4) ) is homogeneous, and that N ˜ e 0 (2) = shows that the problem for N i1

i1

i1

e 0 ˜ e 0 (4) = 0); let us denote N ˜e 0 α = N ˜ e 0 (1) αe ; then the first row n 0 (N ˜ i1 i1 i1 i1 of this matrix-valued function satisfies the equations

α (1)

e 0 α (1)

˜ i1 d2 n

= 0,

dξ˜12

ξ˜1 ∈ e,

(5.8.34)

and the matching conditions at all nodes: (5.8.30), and

e 0 α (1) d˜ n i1 ∗ ¯ + αe γei1 I1,1 αe mesβe = 0, E γe dξ˜1 e(x )

(5.8.34′ )

0

where γe is a column vector with components γei1 that are the cosines of the angles between e and the axis Oxi1 , and I1,1 is the s × s matrix whose (1, 1)−th element is one while the others are zero. This last condition (5.8.34’) can be represented in the form

e 0 α (1) d˜ n i1 ¯ + γei1 Ge mesβe = 0, (5.8.35) E γe dξ˜1 e(x ) 0

where Ge is the s × s matrix whose (i, j)−th element is γe i γe j . For the frames e 0 α (1) = 0. The boundary layer considered in section 8.2 in [16] we have n ˜ i1 e 0 problems for Ni1 are solvable. We also have

e 0 α (1) d˜ n i2 ¯ γe + γei2 Ge dξ˜1 / mesβe dξ˜1 , hi1 i2 = γe i1 mesβe E dξ˜1 e∩Q e∩Q e

e

which coincides with the algorithm for computing the homogenized operator according to the principle of splitting of the averaged operator in section 8.2 in [16] (proposed in [134]). Sufficient condition are also indicated there for ˆ = L

s

i1 ,i2 =1

hi 1 i2

∂2 ∂xi1 ∂xi2

to be the operator of elasticity theory. Suppose that it is the case. Then, after 0 q 0 q ˜e q , M ˜e q all the matrix-valued functions N i1 ...il i1 ...il , j1 ...js , Ni1 ...il and Mi1 ...il , j1 ...js . are constructed, we can proceed with the construction of V, the formal asymptotic solution of the system of equations (5.8.15), which we seek in the form of a regular series in ε and µ :

5.8.


V =

∞

εp1 µp2 Vp1 p2 (x).

305

(5.8.36)

p1 ,p2 =0

where the coefficients of this series are solutions of the recursive sequence of problems ˆ Vp p = fp p (x). LV (5.8.37) 1 2 1 2 where fp1 p2 (x) are defined in terms of the Vq1 q2 with indices (q1 , q2 ) < (p1 , p2 ) that is, qi ≤ pi , i = 1, 2 and q1 = p1 or q2 = p2 , and f00 = f (x); thus, the above equation coincides with the homogenized one obtained in section 5.3 by means of principle of decomposition of the homogenized operator. It can be proved by induction that the fp1 p2 (x) are T −periodic in x and that (0,T )s fp1 p2 (x)dx = 0; consequently, the following assertion holds Lemma 5.8.4 The sequence of problems (5.8.37) is recursively solvable in class of T −periodic functions Vq1 q2 with zero averages, that is, such that V p 1 p2 (x)dx = 0. (0,T )s We denote by u(J) the truncated (partial) sum of the asymptotic series (5.8.7)-(5.8.9), (5.8.20), (5.8.36), obtained by neglecting the terms of order O(µ−2 εJ + µJ ) (in the H 1 (BT )−norm, BT = Bµε ∩ (0, T )s ). Choosing J sufficiently large, we can show for any K that equation (5.8.1) and condition (5.8.2) are satisfied with remainders of order O(µ−2 εK +ε−1 µK ) (in the L2 (BT )−norm. Hence , using the a priori estimate for problem (5.8.1),(5.8.2) given by Theorem 5.8.1, we obtain the following assertion. Theorem 5.8.2 If there are numbers α1 and α2 such that ε = O(µα1 ) and µ = O(εα2 ) as ε, µ → 0, then u − u ˜(J) H 1 (BT ) = O(µ−2 εJ + µJ ), where u ˜(J) = u(J) − < u(J) >T , u is the solution of problem (5.8.1),(5.8.2), and < u(J) >T = 0, < . >T = .dx. BT

Thus, (5.8.7)-(5.8.9), (5.8.20), (5.8.36) is a formal asymptotic solution of problem (5.8.1),(5.8.2). Remark 5.8.1 From the structure of the formal asymptotic solution (5.8.7)(5.8.9), (5.8.20), (5.8.36) it follows that, as ε, µ → 0, the solution does not always converge in the norm .H 1 (BT ) / mesBT

to a function V00 independent of ε and µ; this convergence happens only if ε/µ → 0. For ε/µ = const or ε/µ → +∞ this kind of convergence, generally speaking, does not occur. However, for any relationship between ε and µ the function V00 satisfies a homogenized equation of the form (5.8.37). The main term of the asymptotic expansion coincides with that of section 5.6 where the functions Ni1 ...il have to be extended to Bεµ as constants in every hyperplane

306


orthogonal to the segments e inside of sections and as constants equal to the value Ni1 ...il (ξ0 ) the value in the nodal domain of a node ξ0 . The asymptotic analysis of the elasticity equation in lattice structures by means of the two-scale convergence technique was developed in [67] and it gives the analogous main term.

Figure 5.8.1. Magnitude of the micro-fluctuations of the displacement field.

5.9. RANDOM COEFFICIENTS ON A LATTICE

5.9

307

Random coefficients on a lattice

The Poisson’s equation with random coefficients set in a lattice type domain of a small measure is considered. Such domains simulate the system of fissures and depend on two small parameters: the ratio of the period of the system to the characteristic size of the domain and the ratio of the width of a fissure to the period of the system. The asymptotic analysis is developed.

5.9.1

The Simplest Lattice. The main result.

Consider the simplest model of lattice-structures: two-dimensional rectangular lattice as in section 5.2. Let G be a domain with the boundary ∂G ∈ C ∞ which is independent of ε and µ. Consider a lattice Bε,µ and for each strip Bkj = {(x1 , x2 ) ∈ IR2 | | x3−j − kε | < εµ/2 }, j = 1, 2 we associate a random-valued constant (2 × 2)−matrix AB j , independent of ε and µ. All AB j are independent in aggregate and have k k the same discrete distribution: P {AB j = A(s) } = ps , k

where

r

ps = 1,

s = 1, ..., r,

A(s) = (A(s) )T > 0.

s=1

(s) aij

Let be the elements of the constant matrices A(s) . Let Π be an intersection of the strips: 1 2 Π = ∪+∞ k,j=−∞ (Bk ∩ Bj ),

and A(0) be a fixed constant matrix such that A(0) = (A(0) )T > 0. We pose A = A(0) on the set Π and A = AB j in each strip Bkj \Π without Π. k Let G be a domain with the boundary ∂G ∈ C ∞ which is independent of ε ¯ Consider the problem with random coefficients and µ, f ∈ C 1 (G). −div ( A grad uε,µ ) = f (x) , f or x ∈ Bε,µ ∩ G

(5.9.1)

( A grad uε,µ , n ) = 0, f or x ∈ ∂Bε,µ ∩ G,

(5.9.2)

¯ε,µ ∩ ∂G. uε,µ = 0, f or x ∈ B

(5.9.3)

Problems (5.9.1)-(5.9.3) simulate a problem of permeability of a fissured rock filled with porous substance , with A being the permeability tensor of the substance in the fissures , uε,µ is a microscopic pressure, and u0 is a macroscopic pressure. As it was just noticed in section 5.2, numerical solution of problems

308


(5.9.1)-(5.9.3), with ε ( ε +

Denote for every Bkj = { | x3−j − kε | < εµ/2 , xj ∈ R}, j = 1, 2 N1 (x, ξ3−j ) = (−a−1 3−j,3−j a3−j,j

∂u0 ∂u0 )(ξ3−j − k), − ∂x3−j ∂xj

then ∇u(1) ε,µ (x) = ∇x u0 + ∇ξ (N N1 (x, ξ3−j )ρ(ξξj )ψ(ξ)) + ε∇x (N N1 (x, ξ3−j )ρ(ξξj )ψ(ξ)) = ∇x u0 + ∇ξ N1 (x, ξ3−j ) + R1 , where R1 = ∇ξ (N N1 (x, ξ3−j ) (ρ(ξξj )ψ(ξ) − 1)) + ε∇x (N N1 (x, ξ3−j )ρ(ξξj )ψ(ξ)), R1 2L2 (Bε,µ ∩G) = O(ε + µ)mes(Bε,µ ∩ G) because the measure of the support of the first term is O(ε + µ)mes(Bε,µ ∩ G), and it is bounded and the second term has a value O(ε). 2 ∂u Denote A˜i (∇u) = r=1 ai,r ∂xr and estimate the integral ∂ϕ dx = I = A˜i (∇x u(1) ) ∂x i Bε,µ ∩G

310

CHAPTER 5. LATTICE STRUCTURES 2

j Bε,µ ∩G

j=1

Ai (∇x u0 + ∇ξ N1 )

∂ϕ dx + δ1 , ∂xi

(5.9.5)

2 j,k j,k ∂u where Ai (∇u) = r=1 ai,r ∂xr , and ai,r are the elements of the random matrices AB j in each strip Bkj , and k

√ √ |δ1 | = O( ε + µ) mes(Bε,µ ∩ G)ϕH 1 (Bε,µ ∩G) .

The integral

Ij =

−

j Bε,µ ∩G

2 i=1

Ai (∇x u0 + ∇ξ N1 )

∂ϕ dx = ∂xi

2 ∂ Ai (∇x u0 + ∇ξ N1 )|ξ=x/ε ϕdx, j Bε,µ ∩G i=1 ∂xi

j because A3−j (∇x u0 + ∇ξ N1 ) = 0 in Bε,µ . Therefore ∂ ˆ Aj (∇x v)ϕdx, Ij = − j Bε,µ ∩G ∂xj

where ∂u0 kj kj −1 kj Aˆj (∇x u0 ) = Aj (∇x u0 + ∇ξ N1 ) = (akj . a3−j,j ) jj − aj,3−j (a3−j,3−j ) ∂xj Let ξ˜ be a random variable with the mathematical expectation Eξ and the variance Dξ˜ then for each real η ∈ (0, 1) the Tchebyshev’s inequality takes place ˜ ˜ ≥ Dξ/η} ≤ η. P {|ξ˜ − E ξ| 2 Then for ξ˜ = j=1 Ij we obtain 2 2 2 DIIj /η } ≤ η, P{ | Ij − EIIj | ≥ j=1

where

DIIj

= E +∞

k=−∞

E

j=1

j Bε,µ ∩G

Bkj ∩G

(

(

j=1

2 ∂ ∂ ˆ E Aˆj (∇x v))ϕdx = Aj (∇x v) − ∂xj ∂xj

2 ∂ ∂ ˆ E Aˆj (∇x v))ϕdx . Aj (∇x v) − ∂xj ∂xj

Applying Cauchy-Bunyakowskii-Schwartz inequality we obtain that the last sum is not more than

311

5.9. RANDOM COEFFICIENTS ON A LATTICE

+∞

E

k=−∞

supx∈G

Bkj ∩G

(

∂ ∂ ˆ E Aˆj (∇x v))2 dx Aj (∇x v) − ∂xj ∂xj

Bkj ∩G

ϕ2 dx ≤

2 ∂ ∂ Aˆj (∇x v) − E Aˆj (∇x v)) supj,k mes(Bkj ∩ G)ϕ2L2 (B j ∩G) ≤ ε,µ ∂xj ∂xj C(µε)ϕ2L2 (Bε,µ ∩G) .

Here C is a positive constant independent of ε, µ. Thus P {|

2 j=1

Ij −

2 j=1

EIIj | ≥

2C(µε)ϕ2L2 (Bε,µ ∩G) /η} ≤ η.

Taking 2C(µε) √ √ mes(Bε,µ ∩ G)( ε + µ)2(1−δ)

η =

we obtain the inequality

P {|

2 j=1

Ij −

2 j=1

EIIj | ≥

√ √ mes(Bε,µ ∩ G)( ε + µ)1−δ ϕH 1 (Bε,µ ∩G) } ≤

C1 ε √ √ ( ε + µ)2(1−δ)

≤ C1 εδ ,

(5.9.6)

where C1 ≥ 2Cµ / mes(Bε,µ ∩ G). Consider the intersections of squares Qk1 k2 = { | xi − ki ε | < ε/2 , i = j 1, 2}, with the domain Bε,µ ∩ G. We obtain the equality 2 j=1

−

where

EIIj = −

2 j=1

j Bε,µ ∩G

2

k1 ,k2 :Qk1 k2 ⊂G j=1

j Bε,µ ∩Qk1 k2

∂ E Aˆj (∇x u0 )ϕdx = ∂xj

∂ E Aˆj (∇x u0 )ϕdx + δ2 , ∂xj


Introduce notation

ϕk1 k2 =

Bε,µ ∩Qk1 k2

ϕdx

mes(Bε,µ ∩ Qk1 k2 )

,

312


then the following equality is valid: ϕ = ϕk1 k2 + (ϕ − ϕk1 k2 ),

(5.9.7)

and from Lemma 5.2.4 we obtain the estimate ϕ − ϕk1 k2 L2 (Bε,µ ∩Qk1 k2 ) ≤ Cε∇ϕL2 (Bε,µ ∩Qk1 k2 ) , where C does not depend on ε, µ. Therefore 2 j=1

2

EIIj = −

j Bε,µ ∩Qk1 k2

k1 ,k2 :Qk1 k2 ⊂G j=1

∂ E Aˆj (∇x u0 )dx ϕk1 k2 + ∂xj

δ2 + δ3 , where |δ3 | = O(

k1 ,k2 :Qk1 k2 ⊂G

√ O(ε2 µ

mes(Bε,µ ∩ Qk1 k2 )ϕ − ϕk1 k2 L2 (Bε,µ ∩Qk1 k2 ) ) =

k1 ,k2 :Qk1 k2 ⊂G

∇ϕL2 (Bε,µ ∩Qk1 k2 ) .

From Cauchy-Bunyakovskii-Schwartz inequality for sums we obtain then, that , √ ∇ϕ2L2 (Bε,µ ∩Qk k ) ), |δ3 | = O(ε2 µ ε−2 k1 ,k2 :Qk1 k2 ⊂G

1 2

and therefore √ |δ3 | = O(ε µϕH 1 (Bε,µ ∩G) ) = O(ε) mes(Bε,µ ∩ G)ϕH 1 (Bε,µ ∩G) .

Thus

2

EIIj =

j=1

−

k1 ,k2 :Qk1 k2 ⊂G

Bε,µ ∩Qk1 k2

(1/2)

2 ∂ ˆ Aj (∇x u0 )|x1 =k1 ε,x2 =k2 ε dx ϕk1 k2 + ∂x j j=1

δ2 + δ 3 + δ 4 , where |δ4 | = O((ε + µ)

k1 ,k2 :Qk1 k2 ⊂G

mes(Bε,µ ∩ Qk1 k2 )|ϕk1 k2 |) =

(5.9.8)

5.9. RANDOM COEFFICIENTS ON A LATTICE O((ε + µ)

Bε,µ ∩G

313

|ϕ|dx) = O((ε + µ) mes(Bε,µ ∩ G)ϕH 1 (Bε,µ ∩G) ).

Now we use once more the representation (5.9.7) and Taylor’s expansion for ∂ E Aˆj (∇x u0 ) ∂xj

and obtain that 2 j=1

EIIj = −

Bε,µ ∩G

− where

2 ∂ E Aˆj (∇x u0 )ϕdx + δ5 =, ∂x j j=1

(1/2)

ˆ x u0 )ϕdx + δ5 , div(A∇

Bε,µ ∩G


From the homogenized equation (5.9.4) we can change this integral by 2

EIIj =

2 j=1

2 j=1

f ϕdx + δ5 ,

Bε,µ ∩G

j=1

and therefore

EIIj −

EIIj −

Bε,µ ∩G

f ϕdx =

Bε,µ ∩G

(A∇(uεµ ) , ∇ϕ)dx =

√ √ O( ε + µ) mes(Bε,µ ∩ G)ϕH 1 (Bε,µ ∩G) .

Applying now the estimates (5.9.5), (5.9.6) we obtain: |I − Bε,µ ∩G (A∇(uεµ ) , ∇ϕ)dx| √ √ > ( ε + µ)1−δ } ≤ Cεδ , P{ mes(Bε,µ ∩ G)ϕH 1 (Bε,µ ∩G)

where C does not depend on ε, µ, ϕ i.e.

√ √ 1−δ µ) } ≤ Cεδ . P {uε,µ − u(1) ε,µ F L(Bε,µ ∩G) > ( ε +

Applying now Lemma 5.2.2 and the estimate of section 5.2 (1)

√ uε,µ − u0 L2 (Bε,µ ∩G) √ = O( ε + µ) mes(Bε,µ ∩ G)

314


we get the final estimate P{

√ uε,µ − u0 L2 (Bε,µ ∩G) √ > ( ε + µ)1−δ } ≤ cεδ . mes(Bε,µ ∩ G)

So the theorem is proved. The same consideration is developed for more general types of lattice structures, modelling the systems of fissures oriented in a lot of directions, capillary systems with varying width etc. [135]. The Dirichlet’s problem for Poisson’s equation set in non-periodic lattice structure is considered in [92].

5.10


The lattice-like structures (or skeletal structures) were introduced in [134], where the asymptotic analysis was developed and the principle of splitting the homogenized operator was formulated and justified. The complete asymptotic expansion for a conductivity problem was obtained in [136]. Random structures were studied in [135],[153]. Later lattice-like structures were studied by D.Cioranescu and J. Saint Jean Paulin [46-48]. They considered the result of the passage to limit as the small parameters asymptotic analysis of the lattice-like structures tend to zero in different orders. The L-convergence was introduced in [16] and [155]. The complete asymptotic expansion of a solution to the elasticity equations set in a lattice-like structure was constructed in [145],[148] where the essentially different behavior of the solution was detected in case when the ratio µ/ε is small, finite or great. This result confirms the non-commutativity of the consecutive passages to limit in µ and ε for the elasticity equation described in [44] and [62]; D.Cioranescu and J. Saint Jean Paulin study there the different orders of passage to the limit in so called gridworks , tall structures and honeycomb structures depending on three small parameters µ, ε and e (see Figures 5.9.1 and 5.9.2).


315

Figure 5.9.1. A gridwork

Figure 5.9.2. A tall structure and a honeycomb structure. Nowadays the researches still keep their interest to the lattice-like structures (see, for example, [169] where the non-linear elasticity is considered and justified) and also [66], [67] where the ”regularized” (by some artificial potential) elasticity equation in the rectangular lattice is studied by means of the two-scale convergence.

316


Let us discuss now some models close to the lattice-like structures. The equations on a net were considered in [104]; the asymptotic passage in the difference operators (close to the modelling of lattices) was considered in [126]. Recently the network approximation was studied by L.Berlyand and A.G.Kolpakov in [26] for a high contrast dispersed composite with a small inter-particle distance. The error estimate was then proved in terms of the Voronoy metric in [27]. The fractal-like hierarchical lattice constructions were considered in [79]. The structural mechanics study of the lattice structures can be found in [171]. The lattice-like and the honeycomb structures in a weakly conductive matrix were considered in [17]. The typical result for these structures is as follows. Consider the cell problem [16] for a composite material with square (or cubic in three-dimensional case) periodic inclusions depending on two small parameters : the conductivity of inclusions δ and the thickness of the partition between the inclusions µ. In this case we obtain the explicit formula for the using the principle of splitting of main term of the coefficient of conductivity K = δ + µ(s − 1), where s is the dimension of the homogenized operator [16] : K the space (2 or 3). Let us prove the estimate of the error of this formula. The cases of a finite and of a large conductivity of inclusions are also considered. Consider the following cell problem div(K(x)grad(N N1 + x1 )) = 0, x ∈ IRs for the 1-periodic function N1 of the variables x = (x1 , . . . , xs ) with the piecewiseconstant 1-periodic coefficient K, taking the values : δ inside the cube {|xi − 1/2| < 1/2 − µ/2, i = 1, . . . , s}, and 1 on the other part of the unit cube [0, 1]s . The values of K(x) in each point x ∈ IRs are obtained by the periodic extension. The variational formulation of the cell problem has a form : 1 (K(x)∇(N N1 + x1 ), ∇ϕ) = 0, ∀ϕ ∈ Hper , 1 where . is the integral over cube Q = [−1/2, 1/2]s , Hper is the completion of the space of smooth 1-periodic functions with respect to norm H 1 (Q), N1 is 1 sought in class Hper . The effective conductivity is defined as follows:

= K(x)∂(N K N1 + x1 )/∂x1 .

= δ + (s − 1)µ + O(µ(δ + µ)) if µ → 0, δ → 0, Theorem 4.10.1 K = δ + O(µ) if µ → 0, δ = const, = δ + O(δ 2 µ) if µ → 0, δ → +∞, δµ → 0, = δ/(1 + δµ) + O(1) if µ → 0, δ → +∞, δµ = const, = 1/µ + O((δµ2 )−1 + 1) if µ → 0, δ → +∞, δµ → +∞. Proof. 1/2 Denote i the integral −1/2 dxi . In [16] Chapter 4, section 4.2 (and earlier in the non-published paper by N. S. Bakhvalov) the estimate has been obtained : −1 −1 K −1 −1 1 2 3 ≤ K ≤ K2 3 1 .

K K K K


317

For s = 3 we calculate −1 K −1 −1 (1 − µ) + µ)−1 (1 − µ) + µ)(1 − µ) + µ, 1 2 3 = ((δ

and

−1 −1 K2 −1 (1 − µ) + µ)−1 . 3 1 = (((δ(1 − µ) + µ)(1 − µ) + µ)

We see that the right side and left side of this bilateral estimate have the same asymptotic leading term which is declared in the theorem. Therefore K has also the same asymptotic leading term. The estimate in the two-dimensional case is proved in a same way. The two-scale convergence approach on some domains of degenerating measure was applied to the ”networks”, i.e. the composite materials with some degenerating concentration of a very rigid component [33].

318


5.11

Appendices

5.A1. Appendix 1: the Poincar´ ´ e and the Friedrichs inequalities for lattices

µ µ Let β1(0,d , β3µ , β2(0,d be sets in lemma 4.A2.4 and let β be its union. 1) 2)

1 2 2 Lemma 5.A1.1 In assumptions of lemma 4.A2.4 µs−1 |{u}d1,µ − {u}d2,µ | ≤ 1 i u d x and C10 does not deC10 (∇u)2 d x, where {u}di,µ = µ µ mes βi(0,d β β i(0,di ) i) pend on µ.

Proof. h1 µ h2 µ 1µ 1 2 1 |{u}d1,µ − {u}d2,µ | ≤ |{u}d1,µ − {u}h1,µ | + |{u}1,µ − {u}3,µ | + |{u}3,µ − {u}2,µ |+ h2 µ d2 ′ ′ |{u}2,µ − {u}2,µ ≤ 4 µ1−s C10 β (∇u)2 d x , C10 does not depend on µ (Lem-

mas 4.A2.2, 4.A2.3). Lemma is proved.

Lemma 5.A1.2 Let Bεµ be a lattice and Bµ = {ξ|ǫξ ∈ Bεµ }, let in notations of Theorem 4.A2.1 B = Bµ ∩ QN , QN = (0, N )s , βi , β3j do not depend on ¯ ¯ µ(K) µ s µ where βµ(K) ¯ , β1(0,d1 ) ⊂ B, N , and let β1(0,d 1(0,d1 ) = {x ∈ IR |x − K ∈ β1(0,d1 ) }, 1) ¯ = (K1 , . . . , Ks ), 0 ≤ Ki ≤ N . K Then ¯ 2 d (K)

1 1 µs−1 |{u}d1,µ − {u}1,µ

where

¯ d1 (K) {u}1,µ

1 = mes β µ

1(0,d1 )

¯ µ(K) β 1(0,d

| ≤ C11 N

B

(∇u)2 d x,

u d x, C11 is independent of µ, N .

1)

Proof.

µ µ µ Let β3,j , βiµ1 (0,di1 ) , . . . , β3,j be a path (without repeating) connecting β1(0,d 1 q 1) ¯ µ(K) . Then applying lemma 5.A1.1 to every two consecutive sets βµ with β 1(0,d1 )

il (0,dil )

dl+1 2 µ of the path separated by some set β3j , we have µs−1 |{u}dill,µ − {u}il+1 ,µ | ≤ l C10 (∇u)2 d x with constant C10 from lemma 5.A1.1 µ β i (0,d l

il )

µ ∪β µ ∪β 3j i l

l+1 (0,dil+1 )

which does not depend of µ, N . µ(N ) Here i0 = 1, βiµq+1 (0,di ) = β1(0,d1 ) . Adding these inequalities and applying q+1

319

5.11. APPENDICES Cauchy-Bunyakowskii-Schwartz inequality for sums we have d (N ) 2

1 1 µs−1 |{u}d1,µ − {u}1,µ

≤ 3C10 q

| ≤ µs−1 q

q l=1

d

l+1 2 |{u}dill,µ − {u}il+1 ,µ | ≤

2

B

(∇u) d x ≤ C11 N

B

(∇u)2 d x

. Lemma is proved. Here the length of the path is bounded by const N . Lemma 5.A1.3 Let assumptions of Lemma 5.A1.2 be true and let for any K ∈ {0, 1, . . . , N − 1}s , set BK = B ∩ {ξ ∈ IRs , ξ − K ∈ (0, 1)s } be connected µ(K) and contain β1(0,d1 ) ; let the assumptions of Theorem 4.A2.1 with respect to set BK (instead of set B of Theorem 4.A2.1) be satisfied. Then 1)

⎛

⎝ u2L2 (B) ≤ C12

N

s

µ mesβ1(0,d 1)

µ β 1(0,d

udx

1)

2

+ N s+1

B

⎞

(∇u)2 d x⎠ ;

2) and if B1 = B ∩ {ξ : (ξ2 , . . . , ξs ) ∈ (0, 1)s−1 },be a connected set then ⎞ ⎛

2 N u d x + N2 (∇U )2 d x⎠ , u2L2 (B ) ≤ C12 ⎝ 1 µ 1 mesβµ β B 1(0,d1 )

1(0,d1 )

where C12 does not depend on µ, N .

Proof µ According to Lemmas 4.A2.2 - 4.A2.4 and Lemma 5.A1.2, if β = βi(0,d or i) µ β = β3,j and β ⊂ BK , then

u2L2 (β) ≤ C1

⎛

≤ C2 ⎝

1 µ(K)

1

mesβ

mesβ1,(0,d1 )

β

µ(K)

β1(0,d

udx

2

udx

1)

2

d (K) 2

1 = C3 (µs−1 d1 mesβ1 ({u}1,µ

+ ∇u2L2 (β)

≤

⎞

+ ∇u2L2 (BK ) ⎠ =

) + ∇u2L2 (BK ) ) ≤ d (K) 2

1 1 1 ≤ 2C C3 (µs−1 d1 mesβ1 ({u}d1,µ )2 +µs−1 d1 mes β1 ({u}d1,µ −{u}1,µ

⎛

1 2C C3 ⎝ mesβµ

1(0,d1 )

×

β1µ (0,d1 )

udx

2

+ C11 N

B

2

) +∇u2L2 (BK ) ) ≤

(∇u) d x d1 mesβ1 +

⎞

∇u2L2 (BK ) ⎠,

320


where C3 , C12 do not⎛depend on K, µ. ⎞

2 1 u d x + N (∇u)2 d x⎠ ; So u2L2 (BK ) ≤ C12 ⎝ µ µ mesβ1(0,d β B 1(0,d1 ) 1) ⎞ ⎛

2 s N ⎝ u d x + N s+1 (∇u)2 d x⎠ u2L2 (B) µ ≤ C12 mesβ1(0,d µ β B ) 1 1(0,d ) 1

and

u2L2 (B

1)

⎛

N ≤ C12 ⎝ mesβ µ

µ β 1(0,d

1(0,d1 )

udx 1)

2

+ N2

Proof of Theorem 5.7.1 u d x. Let us take in the last lemma µr = Then,

µ β 1(0,d

1 B

⎞

(∇u)2 d x⎠ .

1)

s+1 (∇u)2 d x. u − µr 2L2 (B) ≤ C4 N B (u − Z)2 d x. Then Let us find now min (u − Z)2 d x. Denote F (Z) = Z β B −2 u d x and min F (Z) = F ( u d x/mesB) = F ′ (Z) = 2(mesB)Z u2 d x− Z B B B 2 1 udx . mesB B So it is less than u − µr 2 2 ≤ C4 N s+1 (∇u)2 dx. Now contracting B 1/ǫ L (B)

B

times we obtain the estimate of Theorem 5.7.1.

Proposition 5.A1.1 Let u ∈ H 1 (G ∩ Bε,µ ), G = (−1, 1)s , u = 0 if x ∈ ∂G ∩ B¯ε,µ . Then uL2 (G∩Bε,µ ) ≤ C∇uL2 (G∩Bε,µ ) where C is independent of ε, µ. Proof. = (−2, 2)s , then for all cylinders ∩ Bε,µ , where G Extend u by zero on G we have u d x = 0. B α+ ⊂ G\G hj

α+ Bhj

Applying the second assertion of Lemma 5.A1.3 to the ”bar” (−2, 2)×(0, ε)s−1 ∩ Bε,µ dilated 1/ε times we obtain that u2L2 (G∩B

ε,µ )

⎛

1 ≤ C12 ⎝ α+ ε mesBhj

= C12

α+ Bhj

∩Bε,µ G

udx

2

+

(∇x u)2 d x,

∩Bε,µ G

⎞

(∇x u)2 d x⎠ =

321

5.11. APPENDICES

, where C12 is a constant of lemma 5.A1.3 part 2. This yields the Proposition. This assertion is valid when u = 0 only on one of the faces of cube G. 5.A2. Appendix 2: the Korn inequality for lattices Consider Bµ = {εx ∈ Bε,µ }, a 1-periodic set. The objective of this appendix is to prove the following theorem. Theorem 5.A2.1 For sufficiently small µ > 0 there exist two positive constants q and C such that, for any 1-periodic vector valued function u defined on 1 Bµ , u ∈ Hper 1 (Bµ ), the inequality holds ∇u2L2 (Bµ ∩(0,1)s ) ≤ CK EBµ ∩(0,1)s (u) where CK = µq C. After the 1ε times contraction, we will obtain the same inequality for Bε,µ : Theorem 5.A2.1’ For sufficiently small µ > 0 there exist two positive constants q and C such that, for any T-periodic function u defined on Bε,µ , u ∈ 1 Hper T (Bε,µ ), the inequality holds

∇u2L2 (Bε,µ ∩(0,T )s ) ≤ CK EBε,µ ∩(0,T )s (u) where CK = µq C. We assume here that for a finite sufficiently small µ this assertion is proved. (The sketch of this prove will be given further in Remark 5.A2.7). Let R be the space of rigid displacements Lemma 5.A2.1 Let G be a domain such that there exists a constant CK (depending on G) such that for any u ∈ (H 1 (G))s , there exists a rigid displacement η ∈ R such that ∇(u − η)2L2 (G) ≤ CK EG (u) (i.e. the Korn inequality holds with a constant CK ). Let G(a,b) be a domain G(a,b) = {x ∈ Rs , ( xa1 , xb2 , . . . , xbs ) ∈ G}, a, b > 0. Then for any (a1 , a2 , b1 , b2 ) such that 0 < a1 ≤ a2 , 0 < b1 ≤ b2 , there exist a constant Cz > 0, depending only on a1 , a2 , b1 , b2 such that for any a ∈ [a1 , a2 ], b ∈ [b1 , b2 ], for any u ∈ (H 1 (G(a,b) ))s , there exists a rigid displacement η ∈ R such that ∇(u − η)2L2 (G(a,b) ) ≤ Cz CK EG (u). ˆ = (au1 , bu2 , . . . , bus ) Proof. Let us make the change x ˆ = ( xa1 , xb2 , . . . , xbs ), u We have ⎧ 2 ∂ui ⎪ ⎨ b ∂xj if i, j = 1, ∂u î ∂ui if i = 1, j = 1 or i = 1, j = 1, ab ∂x = j ⎪ ∂x ˆj ⎩ a2 ∂u1 if i = j = 1 ; ∂x1

let now ηˆ ∈ R, ηˆ =

d−s

αλ ηˆλ , where ηˆ1 (ˆ x) = (−ˆ x2 , x ˆ1 )T , (s = 2) and

λ=1

ηˆ1 (ˆ x) = (−ˆ x2 , x ˆ1 , 0)T , ηˆ2 (ˆ x) = (−ˆ x3 , 0, x ˆ1 )T , ηˆ3 (ˆ x) = (0, −ˆ x3 , x ˆ2 )T , (s = 3)

322


then η(x) =

−1

a

=

5

ηˆ1

x x xs xs xs −2 x1 x2 x2 2 1 −1 = , ,..., , . . . , b ηs , ,..., , b ηˆ2 , ,..., b a b b a b b a b

x

1

x2 , α1 xab1 ) ∈ R (s = 2) (−α 1 ab x2 x2 3 −α1 ab , α1 xab1 , 0 + −α2 xab3 , 0, α2 xab1 + 0, α3 −x b2 , α3 b2 ∈ R (s = 3)

So we have

∇(u−η)2L2 (G(a,b) ) ≤ abs−1 (max(a, b))4 ∇x ( u− η 2L2 (G) ≤ abs−1 (max(a, b))4 1 1 4 4 ) EG(a,b) (u) CK EG ( u) ≤ (max(a, b)) CK (max , a b 4 4 with (max(a, b) max a1 , 1b ) = (max ab , ab ) ≤ max4 ab12 , ab21

Remark 5.A2.1 For homothetic mapping (a = b)C Cz = 1.

Lemma 5.A2.2 Let G1 , G2 be two domains such that G1 ⊂ G2 and such that ∃CE > 0 such that ∀u ∈ (H 1 (G1 ))s , there exists an extension PE u over G2 such that PE u ∈ (H 1 (G2 ))s , PE u = u in G1 , and EG2 (P PE u) ≤ CE EG1 (u). Let G1(a,b) and G2(a,b) be the same mappings of G1 and G2 as in Lemma 5.A2.1. Then for any (a1 , a2 , b1 , b2 ) such that 0 < a1 ≤ a2 , 0 < b1 ≤ b2 , there exists a constant Cz > 0 depending only on a1 , a2 , b1 , b2 (the same that in Lemma 5.A2.1) such that ∀a ∈ [a, a2 ], b ∈ [b1 , b2 ], ∀u ∈ (H 1 (G1(a,b) ))s , there exists an extension PE (a,b) u over G2(a,b) such that PE (a,b) u) ≤ Cz CE EG1(a,b) (u). EG2(a,b) (P Proof. Indeed making the same change as in Lemma 5.A2.1, we consider PE u ˆ, an extension of u from G1 over G2 , and then we make the inverse change and obtain the necessary extension PE (a,b) and EG2(a,b) (P PE (a,b) u) ≤ abs−1 (max(a, b))4 EG2 (P PE u ) ≤ s−1 u) ≤ (max(a, b))4 CE (max a1 , 1b )4 EG1 (a,b) ab (max(a, b))4 CE EG1 (

Thus Lemma 5.A2.2 is proved with the same constant Cz as in Lemma 5.A2.1. Lemma 5.A2.3 Let G1 , G2 be two domains such that G1 ⊂ G2 , and such that there exists a positive constant CH 1 such that ∀u ∈ H 1 (G1 ) there exists an extension u ∈ H 1 (G2 ), u (x) = u(x) in G1 , such that ∇ uL2 (G2 ) ≤ CH 1 ∇uL2 (G1 ) . Assume that ∀u ∈ H 1 (G1 ) there exist a rigid displacement η ∈ R such that ∇(u − η)2H 1 (G1 ) ≤ CK EG1 (u).

5.11. APPENDICES

323

Then there exist the extension of u from G1 over G2 , PE u = u − η + η such that EG2 (u − η + η) ≤ CE EG1 (u). Proof. EG2 (u − η + η) = EG2 (u − η) ≤ ∇(u − η)2L2 (G2 ) ≤ (CH 1 )2 ∇(u − η)2L2 (G1 ) ≤ ≤ (CH 1 )2 CK EG1 (u − η) = (CH 1 )2 CK EG1 (u),

so, CE = (CH 1 )2 CK . Remark 5.A2.2 The assertion of the Lemma 9.A2.3 is valid if H 1 (G1 ) and 1 (G1 ) and H 1 (G2 ) such that ∀u ∈ H 1 (G2 ) H 1 (G2 ) is replaced by subspaces H 1 its restriction on G1 belongs to H (G1 ).

Corollary 5.A2.1 Let β be a bounded star-shaped domain in Rs−1 with a piecewise smooth boundary. Consider the cylinders Cr,µβ = (0, r) × µβ, where µβ = {x ∈ Rs−1 ; µx ∈ β}. Let a1 , a2 , b1 , b2 , positive numbers such that a1 ≤ a2 , b1 ≤ b2 . There exists a unform constant Cz of Lemmas 5.A2.1, 5.A2.2, such that ∀r ∈ [a1 , a2 ], ∀µ ∈ [b1 , b2 ], ∀u ∈ (H 1 (C Cr,µβ ))s , ∃η ∈ R such that

∇(u − η)2L2 (Cr,µβ ) ≤ Cz CK ECr,µβ (u),

(5.A2.1)

CK depends only on β. Proof. Indeed, the Korn inequality for the cylinder C1,β follows from [55]. Then applying Lemma 5.A2.1, we obtain this result. 1 H00 Remark 5.A2.3. Let (H Cr,µβ ))s be the subspace of (H 1 ((C Cr,µβ ))s such (C that its elements vanish at the base x1 = 0. Then the inequality (1) transforms into ∇u2L2 (Cr,µβ ) ≤ Cz CK ECr,µβ (u). Corollary 5.A2.2 Let a1 , a2 , b1 , b2 be positive numbers a1 ≤ a2 , b1 ≤ b2 . There exists a constant Cz of Lemmas 5.A2.1, 5.A2.2, such that ∀r ∈ [a1 , a2 ], ∀µ ∈ [b1 , b2 ], 1) there exists such an extension PE(r,µ) from H 1 (C Cr,µβ ) to H 1 (C Cr,(2µ)β ) such that ECr,µβ (P PE(r,µ) u) ≤ Cz CE ECr,(2µ) β (u), (5.A2.2) where CE depends only on β. 2) there exists such an extension PE (r,µ) from H 1 ((Cr,(2µ)β ∪C(r,2r),µβ ∪C(2r,3r),2µβ )′ ) to H 1 (C C3r,(2µ)β ) such that estimate (5.A2.2) holds true with CE depending only on β; here Cr,µβ = (0, r) × µβ (a cylinder), a bar . is a symbol of closure, A′ is a set of interior points of A.

324


Figure 5.A2.1. The extension. Let 2µ0 be a positive number such that : Bµ is presented as a union of sections S0 and of maximal connected subsets x0 (x0 are the nodal : points) in such a way that the number of sections and of maximal subsets x0 in the unit cube (0, 1)s does not change for all µ ∈ (0, 2µ0 ). (To this end we can chose µ0 from the inequality 2µ0 max diamβ βj j

minx0 mineα1 (x0 ),e2 (x0 ) sin

e; 1 e2 2

µ0 2

hj

hj

0

x0 ,

2

to σ) ; and then we take as usual the continuation Pσ ( u − η) + η where η is α with a a rigid displacement such that the Korn inequality holds true on B hj constant independent of µ and u (Lemma 5.A2.1). −1 When such an extension is constructed for every truncating base σ of 2µ µ0 −1 : : : 2µ µ0 such that from µ0 x0 ,µ to x0 , x0 ,µ we obtain an extension P− of u 2

(P P− u ) ≤ Cµ0

E( u) with the constant Cµ0 independent of µ. At the µ x0 , 0 2 : : second stage we construct an extension of P− u from x0 , µ0 to x0 ,µ0 as in 2 the beginning of the proof and then we make the inverse change of variables x → x. Lemma is proved.

E:

Our goal now is to prove that there exist a uniform constant Cµ0 such that for all µ ∈ [µ0 , 2µ0 ], the Korn inequality holds ∇u2L2 (Bµ ∩(0,1)s ) ≤ Cµ0 EBµ ∩(0,1)s (u)

1 for all 1−periodic functions of Hper (Bµ ). Let us trace the cutting planes separating the sections S0 on 3 parts of the : same height h. For any node x0 denote x0 the nodal domain : x0 reunited with the nearest pieces of sections (the common boundary of x0 and these pieces : also enter in ). For any node x0 make the change x − x0 = µ0 (x − x0 ). x0

2µ

For any segment e denote e the intersection of e with the middle piece of the section corresponding to e. Let x0 be the middle point of e. Make the change µ0 x − x0 = 2µ (x−x0 ). After this change the neighbor pieces of sections : Sx0 (joint : to x0 ) and Sx0 transform into the cylinders Sx0 and Sx0 respectively. Consider now the minimal cylinder Sx0 having the same cross section and including Sx0 and Sx . Its bases are the farthest bases of Sx and Sx . 0

0

0

327

5.11. APPENDICES

Figure 5.A2.4. Extension from Sx0 ∪ Sx0 to Sx0 x0 . x x between the cylinders Sx and Sx0 . Denote Sx0 x0 the open part of C 0 0 0 µ0 µ0 h ), The heights of Sx0 and Sx0 are 2µµ h, while the height of Sx0 x0 is (1 − 2µ µ )(l + 2 where l is the initial distance from x0 to the nearest cutting surface, i.e. l = dist(x0 , x0 ) − h2 : so the height of Sx0 x0 is,

dist(x0 , x0 )(1 −

1 µ0 µ0 ≥ ). ), (1 ≥ 1 − 2 2µ 2µ

Let G = (S x0 ∪ S x0 )′ and let u be a vector-valued function of H 1 (G). = 2µ Make a change u µ0 u. Note that ∇u does not change after the passage from u, x to u , x . Let us prove that there exist such extensions P, PE of u ( x) from Sx0 ∪ Sx0 to 2 ≤ C1 ∇ u2L2 (S ∪S ) , ∀η ∈ R and ∇P u L2 (C Cx0 ,x0 that PE η = η ) x0 ,x0

x0

x0

u), where C1 , C2 > 0 are uniform with respect to PE u ) ≤ C2 ESx ∪Sx ( EC x ,x (P 0 0 0 0 µ ∈ [µ0 , 2µ0 ].

Indeed, let us extend u from Sx0 to Sx0 x0 in such a way that the extension be pair with respect to the common base of the cylinders Sx0 and Sx0 x0 and multiply this extension by a function χµ0 such that depends only on the longitudinal variable xe1 (its axis coincides with the direction of e). This function depends only on µ0 , it is equal to 1 in some neighborhood of the common base of Sx0 and Sx0 x0 and it is equal to 0 at the distance 2l from this base and farther. In the same way we make the extension from Sx0 to Sx0 x0 .

328


For this extension u we have the estimate ∇u L2 (C x

) 0 x0

≤ C 1 ∇ uL2 (Sx

0 ∪Sx0

where C 1 do not depend on µ (it depends on µ0 ). Let now η be such a rigid displacement that ∇(u − η)2L2 ((Sx

0 ∪Sx0 )

′)

≤ C 2 E Sx

0

)

∪Sx0 (u

− η)

C 2 does not depend on µ (Lemma 5.A2.1). Then ∇( u − η( x))2L( S

x0 ∪Sx0 )

≤ C 2 ESx

0 ∪Sx0

( u − η).

Let us construct the above extension u − η for u − η and add the rigid displacement η. We have then u− η) ≤ C 1 ∇( − η) ≤ ∇(u − η)2L2 (C EC x x (u − η+η) = EC x x (u ) 0

0

0

x0 x0

0

2L2 (S

≤ C 1 C 2 ESx ∪Sx ( u). u − η) = C 1 C 2 ESx ∪Sx ( x0 ∪Sx0 ) 0 0 0 0 =u on Sx0 ∪ Sx0 Thus PE u PE u =u − η + η on Sx x . 0

0

Remark 5.A2.4 The extensions P and PE can be constructed in such a man ner that P = PE . Indeed, let us construct first an extension u described above. For this extension u we have the estimates u H 1 (C x

) 0 x0

and

∇u L2 (C x

0 x0 )

uH 1 (Sx ≤ C 1

x ) ∪S 0

0

uL2 (Sx ≤ C 2 ∇

0

x ) ∪S 0

where C 1 , C 2 do not depend on µ (they depend only on µ0 ).

Now construct the solution V of the problem △V + 2graddivV = 0 in Sx0 x0 V =u on ∂ Sx0 ∩ ∂ Sx0 x0 and on ∂ Sx0 ∩ ∂Sx0 x0 s ∂V x x , Bik = (bjl ), bjl = δij δkl + δik δjl + ni = 0 on ∂ Sx0 x0 ∩ ∂ C Bik 0 0 ik ik ∂xk i,k=1

δil δjk , (n1 , . . . , ns ) is a normal vector. We obtain the estimate ∇V L2 (Sx

0 x0 )

L2 (Sx ≤ C 3 ∇u

) 0 x0

where C 3 is uniform with respect to µ ∈ [µ0 , 2µ0 ] (cf Lemma 5.A2.1 and its corollaries, C 3 depends only on β, µ0 , l and h). So

∇V 2L2 (S

x0 x0 )

u2L2 (S ≤ C 4 ∇

x0 ∪Sx0 )

329

5.11. APPENDICES Consider now * v for x ∈ Sx0 ∪ Sx0 Pv = V for x ∈ Sx x 0

0

Let now η be such a rigid displacement that ∇(u − η)2L2 (Sx

0 ∪Sx0

)

≤ C 5 ESx0 ∪Sx0 (u − η)

C 5 does not depend on µ. (Lemma 5.A2.1) Then ∇( u − η( x))2L2 (S

x0 ∪Sx0 )

So

so

∇(V − η( x))2L2 (S

x0 x0 )

ESx

0 x0

≤ C 5 ESx

0

≤ C 4 C 5 ESx

(V − η) ≤ C 4 C 5 ESx

0

0 ∪Sx0

u− x ( ∪S 0

η)

u− x ( ∪S 0

η).

( u − η)

u). and so, ESx x (V ) ≤ C 4 C 5 ESx ∪Sx ( 0 0 0 0 So the lemma is proved. Here the constant C 4 C 5 is uniform with respect to µ ∈ [µ0 , 2µ0 ].

Theorem 5.A2.2 Let µ ∈ [µ0 , 2µ0 ]. Then for any 1-periodic vector-valued 1 function u defined on Bµ , u ∈ Hper 1 (Bµ ), the inequality holds ∇u2L2 (Bµ ∩(0,1)s ) ≤ Cµ0 EBµ ∩(0,1)s (u) where constant Cµ0 is uniform with respect to µ. Proof. Make a change as above and extend the function u ( x) to the cylinders Sx0 x0 as above. Apply now the Korn inequality for Bµ0 /2 (for finite µ0 it is proved as in [55], [129]). Its constant C µ0 /2 depends on µ0

∇ u2L2 (Bµ

0 /2

∩(0,1)s )

u). ≤ C µ0 /2 EBµ0 /2 ∩(0,1)s (

In particular, we have the same estimate for ∇ u2L2 ((B

µ0 /2 ∩(0,1)

s )\

7

x0 ,x0 ∈(0,1)s

x x ) . S 0 0

On the other hand, the extension was made in such a way that EBµ0 /2 ∩(0,1)s ( u) ≤ (2C 1 C 2 + 1)EBµ

0 /2

∩(0,1)s \

7

x0 ,x0 ∈(0,1)s

u) x x ( S 0

where C 1 , C 2 are uniform for µ ∈ [µ0 , 2µ0 ]. Making the inverse change and passing from u , x to u, x we obtain the estimate of the theorem with Cµ0 = C µ0 /2 (2C 1 C 2 + 1).

330


Remark 5.A2.5 In the same way the same estimate can be proved for a finite rod structure when u = 0 on the base of one of cylinders. Now let us prove the main Theorem 5.A2.1’. Let k be chosen in such a way that

min |e|

min |e| e∈B 2k

≥ 2(c0 + 1)µ ≥

min |e| e∈B 2k+1

and let

e∈B

, where c0 is the constant of Definition 5.1.1. 32(c0 + 1) α on 2k parts B α,k (q = 1, . . . , 2k ) of the same Subdivide every cylinder B hj hj,q α in such a way heights by the planes parallel to the bases of the cylinder B

µ0 =

hj

α,k k contains x01 , where x00 and x01 are the ends α,k contains x00 , B that B hj,1 hj,2 k αk αk of the segment eα h , and Bhj,q has a common base with Bhj,q+1 , q = 1, . . . , 2 −1.

Let Πkx0 be the connected part of Bµ cut from Bµ by the nearest cutting planes 1 (Bµ ) to the nodal point x0 . We will make now k − 3 extensions of u ∈ Hper from Bµ to B2µ from B2µ to B22 µ , . . . , and finally from B2k−4 µ to B2k−3 µ in 1 such a way that the extensions u l from B2l−1 µ to B2l µ belong to Hper (B2l µ ) and EB2l µ ( ul ) ≤ C EB2l−1 µ ( ul−1 ) with a constant C independent of µ and of l. Let us describe the construction for a passage from B2l−1 µ to B2l µ (l ≤ k − 4). α (corresponding to the value of small parameAt this stage the cylinders B hj ters that is 2l−1 µ) are divided by the planes parallel to the bases on 2k−(l−1) α,k−(l−1) for this parts as well as the notation parts. We will keep the notation B hj,q k−(l−1)

for the truncated part corresponding to the nodal point x0 that is a Πx0 homothetic extension of Πkx0 in 2l−1 times (x0 is a homothety center). First we k−(l−1) with the coefficient l−1 to Πk−l extend u x0 that is a homothetic image of Πx0 of homothety 2 and the center x0 . This extension is made as follows. Denote k−(l−1) k−(l−1) α,k−(l−1) the part of B2l−1 µ containing Πx0 and all the parts B Π x0 hj,g k−(l−1)

having the common boundary with Πx0

.

331

5.11. APPENDICES

u l (x) =

*

k−l)

k−(l−1)

Figure 5.A2.5. Sets Πx0 and Πx0 xk−(l−1) u l−1 (x) , x ∈ Π 0 k−(l−1) w l (Y (x)) , x ∈ Πk−l \Π x0

.

x0

x ,µ to where w l is an extension of the function y −→ u l−1 (X(y)) from Π 0 0 k Πx0 ,2µ0 ; here Πx0 ,µ0 and Πx0 ,2µ0 are the homothetic images of Πx0 and Πkx0 µ0 and for a homothety with the center x0 and the coefficients respectively µ 2µ0 ; µ y = Y (x) and x = X(y) are changes of variables such that

Y − x0 =

2µ0 (k−l) 2 µ

(x − x0 )

and respectively

2µ0 (X − x0 ). 2(k−l) µ Lemma 5.A2.4 yields that there exist a constant Cµ0 of the extension of any x ,µ ) from Π x ,µ to Πx ,2µ such that the integral EΠ function of H 1 (Π 0 0 0 0 0 0 x0 ,2µ0 is estimated by the integral EΠ x ,µ multiplied by this constant Cµ0 which does 0 0 not depend on the function. It depends on µ0 only. y − x0 =

l−1 (X(y)), obWe apply this extension procedure to the function y −→ u tain the function y −→ w l (y) and then make a back change and obtain x −→ w l (Y (x)). Evidently (as always for a homothetic change of variables) we have the same inequality for u l−1 and w l ◦Y , i.e. EΠk−l \Πk−(l−1) (w l ◦Y ) ≤ Cµ0 EΠ k−(l−1) x0 x0 x0 ( ul−1 ). This constant Cµ0 does not depend on µ or l.

332


α Now consider the extension procedure for the 2k−(l−1) − 4 parts of Bhj that 9 α,l−1 k−l k−(l−1) were not included into Πx0 , i.e. Bhj,q , q = 3, . . . , 2 − 2. Let us x0

α,l−1 denote Cl−1 the cylinder composed of two parts Bhj,q , q = 4, 5.

α,l−1 For Cl−1 as well as for every Bhj,q , q is odd, 7 ≤ q ≤ 2k−(l−1) − 3, make a α

eh ′ eαh )′ where X eαh is a local coordinate system change of variables (X ) = 2(X eαh = α(X − ǫh), and (X eαh )′ corresponding to the segment eα h that is as above X lhα number 2, . . . , s. are the components of X α,l−1 α,l−1 Denote Cl and Bhj,q the maps of Cl and Bhj,q with respect to this change. α,l−1 α,l−1 with the conMake an extension from Cl to Cl as well as from Bhj,q to B hj,q stant Cµ0 such that the extended function has the estimate of EC l and EB α,l−1 hj,q by ECl and EB α,l−1 respectively multiplied by this constant Cµ0 . Such an eshj,q

xk−(l−1) timate can be obtained in a way that is similar to the extension from Π 0 k−l to Πx0 .

2µ0 2(k−l) µ with the center in the middle point of the height of the cylinder (the height that is a part of the segment lhα ). Applying corollary 5.A2.2 (part 1) we obtain such an extension in dilated variables. The back change leaves the same constant for the passage from k−(l−1) Πx0 to Πk−l that is uniform with respect to µ and l. Then for every x0 α,l−1 triplet Bhj,q , q = 2r + 1, 2r + 2, 2r + 3, 2 ≤ r ≤ 2k−(l−1)−1 − 2 as well as

Indeed, we make the homothetic change of variables of a coefficient

α,l−1 for Bhj,q q = 2, 3, 4 we make as above with help of corollary 2 (part 2) an extension from α,l−1 ∪ B α,l−1 ∪ B α,l−1 )′ ) H 1 ((B hj,2r+1 hj,2r+2 hj,2r+3

to

as well as from to

”conserving” the E.

α,l−1 ∪ B α,l−1 ∪ B α,l−1 )′ ) H 1 ((B hj,2r+1 hj,2r+2 hj,2r+3 α,l−1 ∪ B α,l−1 ∪ B α,l−1 )′ ) H 1 ((B hj,2 hj,3 hj,4

α,l−1 ∪ B α,l−1 ∪ B α,l−1 )′ ) H 1 ((B hj,2 hj,3 hj,4

333

5.11. APPENDICES

k−(l−1)

k−l)

Figure 5.A2.6. Extension from Πx0 to Πx0 . 4x to Π3x and then to B2k−3 µ and then At the step k − 4 we extend u from Π 0 0 finally we obtain the extension to B2k−3 µ , where µ0 ≤ 2k−3 µ ≤ 2µ0 . Thus we obtain an extension Pk from Bµ to B2k−3 µ such that EB2k−3 µ ∩(0,1)s (u) ≤ C0 C k EBµ ∩(0,1)s (u), where C does not depend on µ. According to Theorem 5.A2.1 we have∇u2L2(B Pk u2L2(B k−3 ∩(0,1)s ) ≤ Bµ ∩(0,1)s) ≤∇P 2

µ

Cµ0 E(B2k−3 µ ∩(0,1)s ) (P Pk u) ≤ C0 C k Cµ0 EBµ ∩(0,1)s (u) and the Korn inequality is obtained with a constant of order C k = ek lnC , where k = O(|ln µ|) (µ → 0), i.e. k ≤ C|ln µ|, so C k ≤ µ−q , where q does not depend on µ. Thus the theorem on the Korn inequality with a polynomial dependency of the constant on µ is proved.

Remark 5.A2.6 In the same way we can prove the result for H 1 (Bµ ), where Bµ is a finite rod structure and a function vanishes on some part of ∂Bµ that is a base of cylinders constituting Bµ . We apply the same idea of the extension of a vector-valued function vanishing on ∂1 Bµ from Bµ to B2k−3 µ (vanishing on ∂1 Bµ that is the 2k−3 times dilated base ∂1 Bµ ).

1 Remark 5.A2.7 Let us prove the Korn inequality for HperN (Bµ ) for a finite sufficiently small µ > 0, for any N ∈ N with a constant independent of N . To

334


this end we construct an extension from Bµ to Rs in s + 1 steps (”conserving” the energy with a constant independent of N ) Denote Qi1 ...is the unit cube with the center in the point (i1 , . . . , is ) ∈ Ns , Qi1 ...is = {x ∈ Rs , | xj − ij |< 12 , j = 1, . . . , s} ; denote Bia1 ...is the ball of a radius equal center (i1 , . . . , is ) where (2i1 , . . . , 2is ) ∈ Ns , Bia1 ...is = s to a with the s 2 {x ∈ R , j=1 (xj − ij ) < (a)2 }. In case s = 3 consider the cylinders

Cik,a = {x ∈ Rs , 1 ...ik−1 ik+1 ...is

j =k

(xj − ij )2 < a2 },

(2i1 , . . . , 2ik−1 , 2ik+1 , . . . , 2is ) ∈ Ns−1 ; Denote Ba =

9

a B(i , 1 +1/2,...,is +1/2)

i1 ,...,in ∈N

Ca =

s 9

k=1

9

k,a C(i 1 +1/2,...,is +1/2)

i1 ,...,ik−1 ,ik+1 ,...is ∈N

In case s = 2, N even, at the first step we extend the function from Bµ to 9 Qi1 i2 \B a (i1 ,i2 ):i1 +i2

odd

for some finite a < 41 ; at the second step we extend the function further to R2 \B a ; and finally further to R2 . Mention that these first step extensions are made independently for every Qi1 i2 \B a by the same extension operator P1 ; the second step extensions are made independently to every Qi1 i2 \B a with even i1 +i2 from the neighboring Qj1 j2 \B a with odd j1 +j2 (and with the same extension operator P2 ). The third step extensions are made to every Bia1 +1/2,i2 +1/2 from the neighboring Qj1 j2 \B a independently by the same extension operator P3 . The operators P1 , P2 , P3 are those of lemma 5.A2.3 or as in [129].

335

5.11. APPENDICES

2 Figure 5.A2.7. The three steps of extension from 9Bµ to IR . Qi1 i2 \B a If N is odd then the first step set should be i

i

(i1 ,i2 ):i1 +i2 +[ N1 ]+[ N2 ] odd

with a special extension operators for the squares Qi1 i2 with i1 or i2 multiples of N . In case s = 3 (N even) at the first step we extend the function from Bµ to 9 9 Qi1 i2 i3 \(C b B a ) (i1 ,i2 ,i3 )∈N3 :i1 +i2 +i3

odd

7 for some 0 < b < a < 41 ; further to R3 \(C b B a ) ; then to R3 \B a and finally to R3 . Every step of extension (from G1 to G2 ) has been made in such a way that the extension operator P satisfies the estimates EG2 ∩(0,N )s (u) ≤ CEG1 ∩(0,N )s (u), (C is independent of N ), and therefore finally we have constructed an N −periodic extension from Bµ to Rs such that, E(0,N )s (P u) ≤ CEBµ∩(0,N )s (u); so ∇u2L2 (Bµ 2 ∩(0,N )s ) ≤ ∇(P u)L2 (Bµ ∩(0,N )s ) ≤ CK E((0,N )s ) (u) ≤ CK CEBµ ∩(0,N )s (u),where CK is the Korn constant for a periodic cube and it does not depend on N .

Chapter 6

The Multi-Scale Domain Decomposition The method of asymptotic partial decomposition of domain (MAPDD) introduced in [154] is applied here below to partial differential equations, set in rod structures, i.e. in some connected unions of thin cylinders, described in Chapter 4. This method is based on the information about the structure of the asymptotic solution in different parts of such complicated domain. The principal idea of the method is to extract the subdomain of singular behavior of the solution and to reduce dimension of the problem in the subdomain of regular behavior of the solution. The special interface conditions are set on the common boundary of these partially decomposed subdomains. This approach can be easily implemented in the frame of one of modifications of finite element method. Thus this method makes a dimensional (or scale) zoom of some parts of the domain, where the solution has a singular behavior. Rod structures (as it was defined in Chapter 4) are finite connected unions of thin cylinders. The direct numerical solution of partial derivative equations in such domains is very expensive because a complicated geometry demands a large number of nodes in the grid. Another decision is to reduce the dimension and pass to locally one-dimensional model: i.e. each rod is one-dimensional with some junction conditions in the ends. This approach was represented in by sections 4.2, 4.3 but it ignores the boundary layers in neighborhoods of the ends of the rods, while it is well known that these boundary layers are sometimes very important from the point of view of calculation of gradients of solutions: the failure of the rod structure often begins from these neighborhoods because of great concentration of stresses. These stresses can be calculated by constructing of the asymptotic expansion of the solution as in sections 4.7, 4.8, but it is not too easy. Therefore in the present chapter we propose a hybrid method which uses a combined 3D-1D models: it is three-dimensional in the boundary layer domain and it is one-dimensional outside of the boundary layer domain. We cut the

337

338

CHAPTER 6. THE MULTI-SCALE DOMAIN DECOMPOSITION

rods at some distance from the ends of the rods, we keep the dimension three in the neighborhood of the ends and we reduce dimension on the truncated (main) part of rods. Of course the most important question is: what are the interface conditions between 3D and 1D parts? Here below we formulate two approaches of construction of such hybrid models and justify the closeness of the partially decomposed model and initial model. The main principles of construction of such conditions are as follows: i) the asymptotic expansion of the exact solution of the initial problem should satisfy these interface conditions with great accuracy; ii) the hybrid 3D-1D problem with the interface conditions (i.e. partially decomposed problem) should be well posed, i.e. it should have the unique solution and it should be stable with respect to small perturbations in the right hand side. For example, consider the structure at Figure 6.0.1. It is a ”thin domain”, where the partial derivative equations are set with some boundary conditions. The structural mechanics approach reduces it to a completely one-dimensional object (Figure 6.0.2) with some ordinary differential equations along the segments and some junction conditions in the nodes. Although this passage can be justified asymptotically as the small parameter µ tends to zero, the twodimensional information in the neighborhoods of junctions will be completely lost. This information is nevertheless important in the fracture analysis because fractures often start from these domains where the stresses are locally concentrated. From this point of view the method of asymptotic partial decomposition of domain preferable because it passes to a one-dimensional model in the main part of the domain and it keeps the information in some small neighborhoods of the diameter of order εln(ε). Some special interface conditions are set on the contact surfaces between one-dimensional and two-dimensional parts. So, this method can be interpreted as a multi-scale model with the zoom in the neighborhoods of the junctions (Figure 6.0.3). In section 6.1 the differential version of the method of partial asymptotic domain decomposition (MAPDD) is described on some model examples; the general description is given. The variational version is given in section 6.2. The general scheme is discussed; the main theorem about the estimate of the difference between the exact solution and the solution to the partially decomposed problem is proved. The main theorem is then applied to the modelling of thin structures (the finite rod structures). These structures simulate, in particular, the mechanical behavior of the human or animal blood circulatory system [172]. The dimension reduction in such a modelling is a natural approach, although the full-dimensional models have to be kept in the neighborhoods of the bifurcations or junctions. So the MAPDD gives the asymptotically exact answer what should be the correct interface conditions. The flows in such structures are discussed in section 6.3. Other applications (such as an extrusion process) are as well discussed in section 6.3. In section 6.4 the MAPDD is applied to the homogenization problems. The classical homogenization problem [12],[22],[49],[177] of the Dirichlet problem

339 x div(A( )graduε ) = f (x), x ∈ G, uε |∂G = 0, ε is considered; here A(ξ) is an s×s positive definite symmetric matrix 1−periodic in ξ, and ε is a small parameter. Although the asymptotic behavior of the asymptotic solution ”inside” (i.e. at some distance from the boundary) is well studied, the boundary layers are still remain an open problem in the homogenization. And these boundary layers are of great importance for the flux description, for the fracture analysis etc. So, it would be natural to homogenize the problem inside of G at some distance from the boundary keeping the initial formulation in some thin boundary strip. The implementation of this idea arises the question about the asymptotically correct interface conditions between the homogenized and non-homogenized subdomains. We study this question in a model situation when G is a thick layer in IRs , and the asymptotic expansion of the solution is known (it was constructed in [133],[16]). The MAPDD is implemented here below as the variational formulation of the initial problem on the subspace of functions having the form of the asymptotic approximation of order K at some distance from the boundary. Some small corrections transform the MAPDD into the partial homogenization method, prescribing some special interface conditions coupling the homogenized and the non-homogenized parts. Theorem 6.4.1 justifies the partial homogenization by estimating the difference between the asymptotic solution and the solution of the partially homogenized problem.

Figure 6.0.1. A finite rod structure.

340


Figure 6.0.2. The structural mechanics approach: passage to the limit as µ tends to zero (the dimensional reduction).

Figure 6.0.3. Asymptotic partial decomposition of the domain: a hybrid 2D - 3D multi-scale model with the two- dimensional zoom. Some similar hybrid models appeared earlier in mechanics and computations. They are based on some heuristic approaches (see for example, a numerical simulation of shallow water equation in the system ”lake-river” [30]). The open questions were: what is the relation between the original completely 3D model

6.1.

DIFFERENTIAL VERSION

341

and the hybrid model? what is the error? what are the mathematical principles of construction of precise hybrid models, especially the precise interface conditions? Below we shall analyze these questions for hybrid models of conductivity and elasticity equations stated in rod structures. We consider below two versions of the method of asymptotic partial decomposition of domain. The first version is ”differential”, i.e. we work with the differential formulation of the initial problem , we obtain the 1D differential equation in the reduced part of the rod structure and we add the differential interface conditions on the boundary between 3D and 1D parts (as in [26]). Of course we can pass to a variational formulation of the partially decomposed problem but it is generated by a differential one. The second version is a direct variational approach, when the 3D integral identity for the original problem is restated for a special subspace of functions having a form of the ansatz of the asymptotic solution in the regular thin part of the rod structure. This approach is similar to I.Babuska’s idea of the projection with the dimensional reduction, but the partial asymptotic decomposition keeps the initial functional set in the boundary layer domain In this chapter we always use the notation ε for a small parameter, so sometimes it stands for the parameter denoted µ in chapters 1-4.

6.1

Differential version

6.1.1

General description of the differential version

Consider the abstract problem Λ ε u ε = Fε , which depends on a small parameter ε. Here Λε is a linear operator, Fε is a known right hand side and uε is the unknown solution. Let uε , Fε be the functions of the vector valued variable x = (x1 , ..., xs ) varying in the domain Gε . Suppose that the asymptotic behavior of the solution is essentially different in some parts of the domain Gε , i.e. it is a ”regular asymptotic expansion” in a subdomain Gε1 of the domain Gε and it is ”asymptotically singular function” in the subdomain Gε \Gε1 . For example it depends only on the variable x1 in Gε , and in Gε \Gε1 it has the form of the boundary layer (as it will be in subsection 1.1.2), fig.6.1.1. Usually the part Gε \Gε1 is much smaller than the part Gε1 . Then it is natural to decompose the initial problem to a simplified equation in the part Gε1 of Gε (for example, by dimensional reduction as it will be shown in section 1.2) and to the original equation stated in the ”small” reduced part Gε \Gε1 . Of course we should add the conditions on the interface of two parts Gε1 and Gε \Gε1 . The interface conditions should be ”correct”, i.e. the asymptotic solution has to satisfy these conditions, and the whole problem containing the simplified equation set in Gε1 , the original equation posed in the ”small” subdomain Gε \Gε1 and the interface conditions

342


has to be well posed. Let us develop now this idea. Let uε be the solution of some problem depending on a small parameter ε stated in the domain Gε ⊂ IRs ; let the structure of the asymptotic expansion of the solution is known in the ”regular” subdomain Gε1 of the domain Gε , i.e. uε ∼

∞

l=l0

x εl (vl (x) + ul (x, )), ε

(6.1.1)

where ul are bounded and defined identically by the ”previous” functions vl0 , . . . , vl−1 , ul0 , . . . , ul−1 . Assume that we know the relation of ul on these ”previous” functions but we do not know explicitly vl0 , . . . , vl−1 . Suppose that the function uε is expanded asymptotically in the subdomain Gε \Gε1 as follows : ∞

uε ∼

εl Ulε ,

(6.1.2)

l=l0

where Ulε are also bounded but not known. Let each pair of partial sums N (K)

(v (K) , U (K) ) = (

N (K)

εl vl ,

l=l0

be a solution of the problem

εl Ulε )

l=l0

Lε (v (K) , U (K) ) = Fε(K) + O(εK ), H1ε

H2ε

(6.1.3)

H1ε , H2ε

where the linear operator Lε : → is known and are the Hilbert (K) spaces and Fε is an O(εK ) approximation of Fε . Assume that for each F ∈ H2ε there exists a unique solution of the problem of type (6.1.3) Lε V = F and assume that the a priori estimate (stability condition) holds true V H1ε ≤ Cε−r F H2ε where the constants C and r do not depend on ε. ¯ (K) ) of each problem Then clearly the solution (¯ v (K) , U ¯ (K) ) = F (K) Lε (¯(K) , U ε is close to the pair (v

(K)

,U

(K)

(6.1.4)

), i.e.

¯ (K) )H ε = O(εK−r ). (v (K) , U (K) ) − (¯(K) , U 1 Thus the calculation of the terms of the asymptotic expansions (6.1.1) and (6.1.2) up to O(εK−r ) is reduced to the solution of the problem (6.1.4).

6.1.


343

One can build some algorithms based on this idea if the structure of the asymptotic expansions is known . We should only state such a problem (6.1.4) a) which is uniquely solvable and stable and b) for which the partial sums of the asymptotic expansions v (K) and U (K) are the solutions.

6.1.2

Model example

Consider the Poisson equation ∆uε = f (x1 ),

(6.1.5)

2

with the right hand side from L (0, 1), stated in the thin rectangle ε ε Gε = (0, 1) × (− , ) (ε 0) :

Uε ∼

∞ l=0

x εl (vl (x1 ) + Ul ( )). ε

Here v0 (x1 ) is the solution of the boundary value problem 5 ′′ v0 = f, x1 ∈ (0, 1), v0 (0) = 0, v0 (1) = 0;

(6.1.7)

(6.1.8)

Ul (ξ) are the boundary layer terms, i.e. the exponentially decaying (as ξ1 → +∞) solutions of the chain of problems (l = 1, 2, . . .) ⎧ ∆U Ul = 0, ξ ∈ Ω = (0, +∞) × (− 12 , 12 ), ⎪ ⎪ ⎨ Ul = cl for ξ1 = 0, ξ2 ∈ (− 14 , 14 ), (6.1.9) ∂Ul 1 ⎪ ∂ξ2 = 0 for ξ2 = ± 2 , ⎪ ∂U ⎩ 1 ′ l ∂ξ1 = −vl−1 (0) for ξ1 = 0, |ξ2 | ≥ 4 ,

where cl are the constants defined by the condition Ul → 0 as ξ1 → +∞; vl (x1 ) are the solutions of the chain of problems (l = 1, 2, . . .) 5 ′′ vl = 0, x1 ∈ (0, 1), (6.1.10) vl (0) = −cl , vl (1) = 0

344


i.e. vl (x1 ) = cl (x1 − 1); and U0 = 0. Let us state the partially decomposed problem for function v K (x1 ) =

K

εl vl (x1 )

l=0

defined in the domain (δ, 1) (δ 0). Then the following two relations hold true on γδ up to O(εJ )

U (K) (δ, x2 ) = v (K) (δ), 1 ε

ε/2

−ε/2

′ ∂U (K) (δ, x2 )dx2 = v (K) (δ). ∂x1

(6.1.11)

Moreover, for any J there exist J1 ∈ IR such that if δ = J1 ε |lnε| then δ x2 Ul ( , ) = O(εJ ) ε ε

and therefore (6.1.11) holds true up to O(εJ ). Below we take J = K. The information about the structure of asymptotic solution (6.1.7),(6.1.8), (6.1.10) and the relation (6.1.11), true for this asymptotic solution, gives us the conjecture of construction of partially decomposed problem: we keep 3D problem in the neighborhood of end x1 = 0 ε ε ∆U = f (x1 ), x ∈ (0, δ) × (− , ), 2 2 ∂U = 0, x ∈ γ3 ∩ {x1 ≤ δ} ∂n U = 0, x ∈ γ2 ,

we reduce the dimension at the distance δ = O(ε|lnε|) of the end x1 = 0 : v ′′ = f (x1 ), x ∈ (δ, 1), v(1) = 0, and we state the interface conditions induced by (6.1.11) U (δ, x2 ) = v(δ),

1 ε

ε/2

−ε/2

∂U (δ, x2 )dx2 = v ′ (δ), ∂x1

(6.1.11′ )

6.1.

345


i.e. the partially decomposed problem is v ′′ = f (x1 ), x ∈ (δ, 1), v(1) = 0, ε ε ∆U = f (x1 ), x ∈ (0, δ) × (− , ), 2 2 ∂U = 0, x ∈ γ3 ∩ {x1 ≤ δ} ∂n U = 0, x ∈ γ2

U (δ, x2 ) = v(δ),

1 ε

ε/2

−ε/2

(6.1.12)

∂U (δ, x2 )dx2 = v ′ (δ). ∂x1

Figure 6.1.1. The ”cutting” procedure in the method of partial asymptotic decomposition of domain. The pair (v (K) , U (K) ) satisfies (6.1.12) exactly with the exception of the condition ∂U ε = 0 for x1 = 0, |x2 | ≥ , 4 ∂n

which is satisfied up to the term O(εK ), i.e. ∂U (K) ′ = −εK vK (0), ∂x1

and with exception of the conditions (6.1.11 ’) satisfied up to O(εK ). So the pair (v (K) , U (K) ) satisfies the problem of the type (6.1.4).

(6.1.13)

346


The variational formulation of the problem (6.1.12) is as follows. Let H 1 ((0, δ)×(− 2ε , 2ε ), (δ, 1)) be the space of pairs of functions (R(x1 , x2 ), ρ(x1 ))such that

ε ε R ∈ H 1 ((0, δ) × (− , )), ρ ∈ H 1 ((δ, 1)), R(δ, x2 ) = ρ(δ), R|γ2 = 0, ρ(1) = 0. 2 2

We seek such an element (U, v) ∈ H 1 ((0, δ) × (− 2ε , 2ε ), (δ, 1)), that for any

ε ε (R, ρ) ∈ H 1 ((0, δ) × (− , ), (δ, 1)), 2 2 1 δ 2ε (∇U, ∇R)dx1 dx2 − εv ′ ρ′ dx1 = − 0

=

− ε2

0

δ

δ

ε 2

− 2ε

f Rdx1 dx2 +

1

εf ρdx1 .

(6.1.14)

δ

Remark 6.1.1 Problem (6.1.12) is ”better” than (6.1.5), (6.1.6) because of the reducing of the two dimensional part of the problem from Gε to G1ε . The part Gε \G1ε is replaced by a one-dimensional segment. The problem set in G1ε is responsible for the boundary layer and it is coupled with the one-dimensional regular problem stated in (δ, 1). It contains complete information on the asymptotic expansion of the solution of the problem (6.1.5), (6.1.6). Now this coupled 2D - 1D problem (6.1.12) could be solved numerically by the finite element method. To this end we consider the standard triangle partition of the two dimensional part and a uniform partition of the one dimensional part . Enumerate all nodes. We introduce then the standard piecewise linear base hat functions (independently in each of two parts; the hat function number n is equal to 1 in one of the nodes of the grid and it vanishes in all other nodes; these hat functions are defined for all nodes with exception of the nodes belonging to the interface line γδ . Formally, this set of base functions corresponds to two independent sets of standard hat functions defined in 2D part and in 1D part with the Dirichlet condition on γδ . Then we complete the union of these two sets of 2D and 1D hat functions by one special hybrid hat function (”super-element”) which is defined on the union Gε ∪ (δ, 1) and it is partially 2D and partially 1D hat function: it is piecewise linear, it is equal to 1 on γδ (i.e. for all nodes of γδ ) and it is equal to zero in all other nodes. Finally, we extend all standard 2D hat functions by zero on the 1D part, we extend all standard 1D hat functions by zero on the 2D part, and then use all hat functions (the hybrid ”super-element” included) to form a base of finite element subspace Hf1e and we state the problem (6.1.14) on this subspace, i.e. (R, ρ) ∈ Hf1e . We developed the direct numerical study of the model 2D problem (6.1.5),(6.1.6) by finite element method and the numerical study of the partially decomposed problem (6.1.14). The comparison of results of these numerical experiments shows the excellent precision of the partially decomposed problem with respect to the original 2D problem.

6.1.


347

Figure 6.1.2. Two dimensional triangulation for direct finite element method and for a partially decomposed domain.

Figure 6.1.3. The finite element implementation: a) a 1D basis ”hat”-function; b) a 2D basis ”hat”-function; c) the hybrid 2D-1D basis ”hat”-function. The same numerical experiment for a one bundle rod structure (see definition below in section 6.1.3) confirms the closeness of the original and the hybrid models.

348


Now to justify the passage from the problem (6.1.5), (6.1.6) to the problem (6.1.12), we should prove the existence and uniqueness of the solution of the problem (6.1.12) (which is not standard) and prove its stability with respect to perturbations in the right hand sides of the conditions (6.1.13) and (6.1.11’). Note that the question of stability could be reduced to the stability of the condition (6.1.11’) only. Indeed let U c be the solution of the following boundary layer problem : ⎧ ∆U c = 0 for ξ ∈ Ω, ⎪ ⎪ c ⎪ ⎪ ξ1 = 0, ξ2 ∈ (− 14 , 14 ), ⎨ U = 0 for 1 ∂U c ∂ξ2 = 0 for ξ2 = ± 2 , c ⎪ 1 ∂U ⎪ ⎪ ⎪ ∂ξ1 = 1 for ξ1 = 0, |ξ2 | ≥ 4 , ⎩ U c → const as ξ1 → +∞.

Add the corrector εK v(0)U c ( xε ) to U (K) in the domain G1ε = (0, δ) ×

(− 2ε , 2ε ).

√ Note that the norm H 1 (G1ε ) of this corrector is O(εK−1 εδ) and that the corrected solution satisfies to all conditions (6.1.12) exactly and to conditions (6.1.11’) up to the term O(εK ). Thus the justification of the passage from the problem (6.1.5), (6.1.6) to the problem (6.1.12) is reduced to the questions of existence and uniqueness of the solution (6.1.12) and its stability with respect to the perturbations in the conditions (6.1.11’) only. Let us prove now that the problem (6.1.12) is uniquely solvable. Consider the variational formulation (6.1.14) (the other proof is given for classical formulation in [156]). The Poincar´ é-Friedrichs inequality gives:

RL2 ((0,δ)×(− 2ε , 2ε )) ≤ Cδ∇RL2 ((0,δ)×(− ε2 , 2ε )) ,

where C does not depend on δ, ε, and ρL2 ((δ,1)) ≤ Cρ′ L2 ((δ,1)) . So the right hand side functional is bounded by √ C εf L2 ((0,1)) (∇R2L2 ((0,δ)×(− ε , ε )) + ερ′ 2L2 ((δ,1)) )1/2 . 2 2

Applying the Lax-Milgram lemma we obtain the existence and uniqueness of the solution. We obtain also the a priori estimate U 2H 1 ((0,δ)×(− ε , ε )) + εv2H 1 ((δ,1)) ≤ U, v1 = 2 2

√ C1 εf L2 ((0,1)) .

The perturbations in the conditions (6.1.11’) could be implemented to the right hand side by simple subtraction of a pair of functions Ψ ∈ H 1 ((0, δ) × (− 2ε , 2ε )) and ψ ∈ H 1 ((δ, 1)) such that

6.1.


349

1) this pair satisfies the boundary conditions ∂Ψ/∂n = 0, x ∈ γ3 ∩{x1 ≤ δ}, Ψ = 0 on γ2 and ψ(δ) = ψ(1) = 0; 2) the trace of the function Ψ(δ, x2 ) is equal to the given perturbation in the ε/2 ∂Ψ (δ, x2 )dx2 is equal to the first condition (11’) and the value ψ ′ (δ) − 1ε −ε/2 ∂x 1 given perturbation in the second condition (6.1.11’) The structure of the asymptotic solution Uε shows that for any K, there exists K1 ∈ IR such that if δ = K1 ε |lnε| then the perturbations in (6.1.11’) are of order O(εK ) . Therefore again for any K, there exist K1 ∈ IR such that if δ = K1 ε |lnε| then

U Uε(K) − U , v (K) − v1 = O(εK ). On the other hand, U Uε(K) − uε H 1 (B ε ) = O(εK ). So for any K, there exist K1 ∈ IR such that if δ = K1 ε |lnε| then uε − U H 1 ((0,δ)×(− ε2 , 2ε )) = O(εK ),

uε − vH 1 ((δ,1)×(− ε2 , 2ε )) = O(εK ).

6.1.3

Poisson equation in a rod structure

In the same way the partially decomposed problem can be set for the Poisson equation in a rod structure ( the asymptotic expansion of the solution was built in Chapter 4 and [148]). It could be proved that the same estimate holds true. Describe the method in a simple case of a rod structure containing one bundle. Let e1 , ... , en be n closed segments in IRs (s = 2, 3), which have a single common point O (i.e. the origin of the coordinate system), and let it be the common end point of all these segments. Let β1 , ..., βn be n bounded (s-1)dimensional domains in IRs , which belong to n hyper-planes containing the point O. Let βj be orthogonal to ej . Let βjε be the image of βj obtained by a homothetic contraction in 1/ε times with the center O. Denote Bjε the open cylinders with the bases βjε and with the heights ej , denote also βˆjε the second base of each cylinder Bjε and let Oj be the end of the segment ej which belongs to the base βˆjε . Define the bundle of segments ej centered in O as B = ∪nj=1 ej . Denote below O0 = O. Define the one bundle rod structure associated with the bundle B as the set B ε of inner points of the union ∪nj=1 B¯jε . Here the bar means the operation of the closure.

350


Figure 6.1.4. A one bundle rod structure. e e Introduce the local coordinate system Ox10j ...xs0j associated with a segment e0j ej such that the direction of the axis Ox1 coincides with the direction of the e e e segment OOj , i.e. x10j is a longitudinal coordinate. The axes Ox10j , ..., Oxs0j form a cartesian coordinate system. Define some cutting functions χj on IR+ . Let dj ε be the minimal distance from the base of the cylinders Bjε to a parallel cross-section σjε of Bjε such that this cross-section does not contain points of other cylinders (i.e. σjε ∩ e e Biε = ∅, i = j). Then let χj = 0 for x10j ≤ dj and χj = 1 for x10j ≥ dj + 1; 2 we suppose that the function χj ∈ C and it does not depend on ε. e e In the same way we can introduce local coordinate systems Oj x1j0 ...xsj0 ej0 associated with ej such that the direction of the axis Oj x1 coincides with the e e direction of the segment Oj O, and so x10j = |ej | − x1j0 . Consider the Poisson equation ∆uε = f (x), x ∈ B ε

(6.1.15)

with the boundary conditions uε = 0 for x ∈ βˆjε , j = 1, ..., n,

(6.1.16)

∂uε = 0 for x ∈ ∂B ε \ ∪nj=1 βˆjε , ∂n

(6.1.17)

e

e

where f is equal to fj (x10j )χj (x10j /ε) for each ej , fj ∈ C ∞ ((0, |ej |)). The asymptotic analysis of this problem developed in Chapter 4 shows that the boundary layers decay exponentially.

6.1.

351


Cut off the cylinders Bjε at the distance δ = const ε ln (ε) from the nodes by the planes perpendicular to the segments and replace the inner parts of the cylinders by the corresponding parts of the segments ej . We obtain the set B ε,δ . Denote Biε,δ the connected truncated part of B ε , containing the node Oi , and denote eij the part of one segment e = Oi Oj among e1 , ..., en which connects Biε,δ and Bjε,δ ; let Sij be a cross-section of the truncated cylinder corresponding to eij , such that it belongs to ∂Biε,δ . Consider the equations for each Biε,δ : ∆U = f,

(6.1.18)

U = 0, x ∈ βˆjε ,

(6.1.19)

∂U Uε = 0, x ∈ (∂Biε,δ ∩ ∂B ε )\ ∪nj=1 βˆjε , ∂n the equation for each segment ej1 ,j2 : e

vj′′1 ,j2 = f (x1j1 ,j2 ),

(6.1.20)

(6.1.21)

and the interface conditions on each cross-section Sj1 ,j2 e

U = vj1 ,j2 , x1j1 ,j2 = δ; ∂U 1 ds = vj′ 1 ,j2 (δ), mesS Sj1 ,j2 Sj1 ,j2 ∂n

(6.1.22)

(6.1.23)

e

where vj1 ,j2 are unknown functions of the variable x1j1 ,j2 defined on each segment ej1 ,j2 . Of course e

e

vj1 ,j2 (x1j1 ,j2 ) = vj2 ,j1 (|ej1 ,j2 | + 2δ − x1j1 ,j2 ).

(6.1.24)

Let us give the variational formulation for this problem. Let H 1 (B0ε,δ , ..., Bnε,δ , e1 , ..., en ) be the space of ordered collections (U U0 , ..., Un , w1 , ..., wn ), where Ui are the functions from H 1 (Biε,δ ), equal to zero on ∂Biε,δ ∩ ∪nj=1 βˆjε , and w1 , ..., wn are the e functions of the variable x1j associated with the segments e1 , ..., en such that for each Sj1 ,j2 we have Uj1 = wj , (6.1.25) where ej = ej1 ,j2 . The scalar product is defined as ((U U0 , ..., Un , w1 , ..., wn ), (V V0 , ..., Vn , v1 , ..., vn ))P = n s ∂U Ui ∂Φi e s−1 dx + ε mesβ βj vj′ wj′ dx10j . ε,δ ε,δ n ∂x ∂x r r Bi r=1 j=1 ej \∪i=0 Bi

n i=0

352


The norm is defined as 1/2

(U U0 , ..., Un , w1 , ..., wn )P = ((U U0 , ..., Un , w1 , ..., wn ), (U U0 , ..., Un , w1 , ..., wn ))P . Then the variational formulation is as follows: find (U U0 , ..., Un , v1 , ..., vn ) ∈ H 1 (B0ε,δ , ..., Bnε,δ , e1 , ..., en ) such that for any (Φ0 , ..., Φn , w1 , ..., wn ) ∈ H 1 (B0ε,δ , ..., Bnε,δ , e1 , ..., en ) the integral identity holds true:

−

n s ∂U Ui ∂Φi e dx − εs−1 mesβ βj vj′ wj′ dx10j ε,δ ε,δ n ∂x ∂x r r e \∪ B Bi r=1 j i=0 i j=1

n i=0

n

=

Biε,δ

i=0

βj + εs−1 mesβ

(f, Φ)dx

n

e

ε,δ ej \∪n i=1 Bi

j=1

f wj dx10j .

(6.1.26)

The asymptotic analysis of the problem (6.1.15)- (6.1.17) shows that for any K we can find such K1 ≥ K that if δ = K1 ε |lnε| then the truncated sum u(K) of order K of the asymptotic expansion (see section 4.7) adopted to a scalar (K) case and defined on Biε,δ and the truncated sum vj of order K of the regular e0j asymptotic expansion (depending on x1 only) j = 1, ..., n satisfy to the same integral identity with the discrepancy n i=0

∂Biε,δ \∂B ε

εK ρ1 Φi ds +

+ εK+(s−1)

n i=0

n j=1

Biε,δ

εK ρ2 Φi dx e

ε,δ ej \∪n i=1 Bi

r3,j wj′ dx10j .

(6.1.27)

There is also a discrepancy in the condition (6.1.22) of order O(exp(−cδ/ε)), i.e.

(K1 )

u(K1 ) = vj

+ exp(−cδ/ε)r3 ,

x ∈ Sj1 ,j2 ,

(6.1.28)

where c > 0, ρ1 , ρ2 , r3,j are bounded vector valued functions. Moreover the discrepancy exp(−cδ/ε)r3 can be continued from Sj1 ,j2 to Biε,δ in such a way that this extension has an order O(exp(−c1 δ/ε)), c1 > 0 in H 1 . Subtracting this extension from the truncated series u(K) we obtain the integral identity with the final discrepancy of a form (6.1.27). The functional (6.1.27) is bounded in the norm P due to the estimate Φi L2 (B ε,δ ) , εs−1 mesβ βj wj L2 (ej \∪n

ε,δ i=1 Bi )

, Φi L2 (∂B ε,δ \∂B ε )

6.1.

353

DIFFERENTIAL VERSION ≤ CΦ0 , ..., Φn , w1 , ..., wn P

(6.1.29)

with the constant C independent of ε, δ. This estimate is a corollary of the Poincar´ é - Friedrichs inequality for finite rod structures (see Appendix 4.A2). ΦL2 (B ε ) ≤ C∇ΦL2 (B ε )

(6.1.30)

if we apply it to the function Φ defined as Φ(x) =

*

Φi (x), x ∈ Biε,δ , e0j wj (x1 ), x ∈ Bjε \ ∪ni=1

i = 0, ..., n, Biε,δ , j = 1, ..., n.

For Φi L2 (Sj1 ,j2 ) we should also use the standard estimate Φi L2 (Sj1 ,j2 ) ≤ CΦH 1 (B ε,δ

ε,δ n j1 ,j2 \∪i=1 Bi )

,

where Bjε,δ is the cylinder Bjε,δ which corresponds to the surface Sj1 ,j2 (i.e. 1 ,j2 contains this surface). The integral identity (6.1.26) has the unique solution due to the Lax-Milgram lemma. Taking into consideration the discrepancies (6.1.27) we obtain for the difference a following estimate (K)

(u(K) |B ε,δ − U0 , ..., u(K) |Bnε,δ − Un , v1

− v1 , ..., vn(K) − vn )P = O(εK ), (6.1.31) Combining the estimate (6.1.31) with the estimate of the difference between the exact and asymptotic solutions we obtain that for any K there exist such ˆ independent of ε that if δ = Kε|ln(ε)| ˆ K then the estimate holds true 0

U − uH 1 (B ε ) = O(εK ),

(6.1.32)

where U (x) =

*

Ui (x), if x ∈ Biε,δ , e0j vj (x1 ), if x ∈ Bjε \ ∪ni=0

i = 0, ..., n, Biε,δ , j = 1, ..., n,

where the cylinder Bjε corresponds to ej . The estimate (6.1.32) justifies the method of asymptotic partial decomposition of domain (MAPDD). Remark 6.1.2. This construction can be easily generalized to the case of a finite rod structure with m bundles of bars: 1 m B1 = ∪nj=1 ej,1 , ..., Bm = ∪nj=1 ej,m .

354


We suppose that all common points of these bundles are end points of some segments of these bundles. Let the union of all bundles be connected. Consider the rod structure Bαε associated with the bundle Bα and let ε B ε = ∪m α=1 Bα

be connected. Then the same asymptotic decomposition of the domain gives the estimate (6.1.31). We can also add some small domains in the neighborhoods of the nodes, as it was done in section 4.1.

6.2

Variational version

6.2.1

General description of the variational version

Let Hε be a family of Hilbert spaces (depending on a small parameter ε). Consider a variational problem: find uε ∈ Hε such that ∀w ∈ Hε , B(uε , w) = (f, w).

(6.2.1)

Here B(., .) is a bilinear symmetric coercive form and (f, .) is a linear bounded functional with the norm f . We suppose that ∀w ∈ Hε , B(w, w) ≥ c1 εr w2 ,

(6.2.2)

where c1 > 0 and r ≥ 0 do not depend on ε, ∀u, w ∈ Hε , B(u, w) = B(w, u).

(6.2.3)

On the other hand, by definition, we get ∀w ∈ Hε , |(f, w)| ≤ f w.

(6.2.4)

Then it is well known that there exists a unique solution uε of problem (6.2.1). Let Hε,dec be a linear subspace of Hε . Let uaε be an asymptotic solution such that (i) uaε ∈ Hε,dec , and such that: (ii) there exists a functional (ψε , .) ∈ Hε∗ such that its norm ψε ≤ c2 , where c2 does not depend on ε, and ∀w ∈ Hε , B(uaε , w) = (f, w) + εK (ψε , w), where K > r. Subtracting (6.2.1) from (6.2.5) we obtain ∀w ∈ Hε , B(uaε − uε , w) = εK (ψε , w),

(6.2.5)

6.2.

VARIATIONAL VERSION

355

i.e. for w = uaε − uε we obtain c1 εr uaε − uε ≤ εK ψε ≤ εK c2 , i.e. uaε − uε ≤ (c2 /c1 )εK−r , i.e. uaε − uε = O(εK−r ).

(6.2.6)

udε

Let be a solution (its existence is assumed)of the partially decomposed problem, i.e. of the identity (6.2.1) restricted onto the subspace Hε,dec : ∀w ∈ Hε,dec , B(udε , w) = (f, w).

(6.2.7)

We assume the existence and uniqueness of solution udε of problem (6.2.7). We assume that subspace Hε,dec has a more simple structure than Hε , for example the functions of Hε,dec are polynomial on the regular part Gε1 of the domain. Therefore the problem (6.2.7) is in some sense easier than problem (6.2.1). So the variational version is related to a special choice of a simple subspace Hε,dec (i.e. special restriction of the original problem), satisfying the conditions (i), (ii). To justify the variational version let us subtract the identity (6.2.7) from (6.2.5) for any w ∈ Hε,dec . Then we obtain ∀w ∈ Hε,dec , B(uaε − udε , w) = εK (ψε , w), i.e. for w = uaε − udε we obtain c1 εr uaε − udε ≤ εK ψε ≤ εK c2 , i.e. uaε − udε ≤ (c2 /c1 )εK−r , i.e. uaε − udε = O(εK−r ).

(6.2.8)

Comparing the estimates (6.2.6) and (6.2.8) we obtain uε − udε = O(εK−r ).

(6.2.9)

This estimate justifies the method. We can also consider a more general problem, when Bε,t is an arbitrary ( ˜ ε,T × Hε,T to IR satisfying instead of the possibly non-linear) mapping from H above inequality the following relation: ˜ ε,T , ∀w1 , w2 ∈ H

356


sup (Bε,t (w1 , w1 − w2 ) − Bε,t (w2 , w1 − w2 )) ≥ c1 εr w1 − w2 1+α , (6.2.2′ ) T

t∈(0,T )

here α is a positive constant. Now the above proof of estimate (6.2.9) is modified as follows. ˜ ε,T . Let uaε be an asymptotic solution such that Let Hε,dec be a subspace of H (i) uaε ∈ Hε,dec , and such that ∗ (ii) there exists ψε ∈ Hε,T , such that ψε ≤ c2 , where c2 does not depend on ε, and such that for almost all t ∈ (0, T ), ∀w ∈ Hε,T , Bε,t (uaε , w) = (fft , w) + εK (ψε , w),

(6.2.5′ )

where K > r. Subtracting (6.2.1) from (6.2.5’) we get ∀t ∈ (0, T ), ∀w ∈ Hε,T , Bε,t (uaε , w) − Bε,t (uε , w) = εK (ψε , w), i.e. for w = uaε − uε passing to supt∈(0,T ) we have c1 εr uaε − uε Tα ≤ εK ψε T ≤ εK c2 , i.e. uaε − uε T = O(ε

K−r α

).

(6.2.6′ )

Let udε be the solution of the partially decomposed problem, i.e. of the identity restricted onto the subspace Hε,dec : ∀w ∈ Hε,dec , Bε,t (udε , w) = (fft , w),

(6.2.7′ )

As above, we assume that this subspace has a more simple structure than Hε,T . Let us subtract this identity from (6.2.1) written for any w ∈ Hε,dec . Then we obtain ∀t ∈ (0, T ), ∀w ∈ Hε,dec , Bε,t (uaε , w) − Bε,t (udε , w) = εK (ψε , w), i.e. for w = uaε − udε passing to supt∈(0,T ) we obtain K K c1 εr uaε − udε α T ≤ ε ψε T ≤ ε c2 ,

i.e. uaε − udε T = O(ε

K−r α

).

(6.2.8′ )

Comparing estimates (6.2.8’) to (6.2.6’) we get the following assertion. Theorem 6.2.1 The estimate holds uε − udε T = O(ε

K−r α

).

This theorem generalizes estimate (6.2.9) and justifies the MAPDD for the non-linear and non-steady-state case.

6.2.

357

VARIATIONAL VERSION

6.2.2

Model example

Returning to the example (6.1.5),(6.1.6), we can define Hε as a subspace of functions of H 1 ((0, 1) × (− 2ε , 2ε )) vanishing on γ1 ∪ γ2 . Then the identity (6.2.1) is

0

1

ε 2

− 2ε

(∇uε , ∇w)dx1 dx2 = −

0

1

ε 2

− 2ε

f wdx1 dx2 .

(6.2.10)

Consider now the subspace ε ε Hε,dec = {w ∈ Hε , w(x1 , x2 ) = ρ(x1 ) in (δ, 1) × (− , ), ρ ∈ H 1 ((δ, 1))}. 2 2

Evidently, the partially decomposed problem in this second version coincides with the problem (6.1.14) up to the notations: 5 U (x1 , x2 ) if x1 < δ, uε (x1 , x2 ) = v(x1 ) if x1 ≥ δ, w(x1 , x2 ) =

5

R(x1 , x2 ) if x1 < δ, ρ(x1 ) if x1 ≥ δ.

The asymptotic solution uaε should be slightly corrected because in a form (6.1.7’) it does not belong to the subspace Hε,dec (it is not a constant on the rectangle (δ, 1) × (− 2ε , 2ε ) although the error is exponentially small). Therefore we multiply the boundary layer functions Ul at the domain where they are exponentially small by a cutting function χ; i.e. let χ be a smooth function defined on IR such that 5 1 if t < −1, χ(t) = 0 if t ≥ 0,

|χ(t)| ≤ 1, then uaε (x) =

K l=0

x1 − δ x )). εl (vl (x1 ) + Ul ( )χ( ε ε

Clearly δ

uaε − U (K) H 1 (Gε ) = O(e−c3 ε ), c3 > 0

due to exponential decaying of Ul (ξ) as ξ1 → +∞, and this correction gives a discrepancy of the same order in the variational formulation (6.2.10). So for any K we can find K1 such that if δ = K1 ε |ln ε| then δ

e−c3 ε = O(εK ),

358


and the conditions (i) and (ii) of the subsection 6.2.1 are satisfied and we again obtain the estimate uaε − udε H 1 (Gε ) = O(εK ),

(6.2.11)

confirming the result of the previous section 6.1. (Here r = 0 due to the Poincare-Friedrichs ´ inequality for rod structures). Of course the same approach is good for the general rod structures.

6.2.3

Elasticity equations

Consider the elasticity system of equations set in ε ε Gε = (0, 1) × (− , ) 2 2 (below the convention of the summation from 1 to 2 in repeating indices is accepted):

∂uε x ∂ (Arm ) = f ( ), x ∈ Gε , ε ∂xr ∂xm

(6.2.12)

where Arm are constant 2 × 2 matrices with the components akl rm : akl rm = λδrk δlm + µ(δrm δkl + δrl δkm ), λ and µ are the positive constants ; here uε , f are two- dimensional vectorvalued functions, supp f (ξ) belongs to the square (0, 1) × (− 21 , 12 ), and f ∈ L2 . Let us set the following boundary conditions: free lateral boundary and fixed ends:

∂uε = 0, x2 = ±ε/2 ∂xm

(6.2.13)

uε = 0, x1 = 0, 1.

(6.2.14)

A2m

The variational formulation is the integral identity (6.2.1) where ∂uε ∂w T dx, ) Aij B(uε , w) = ( ∂ξξj Gε ∂ξi (f, w) = − f wdx,

(6.2.15)

Gε

Hε is the subspace of vector valued functions of [H 1 (Gε )]2 vanishing on the segments {x1 = 0} and {x1 = 1} . The complete asymptotic expansion of the solution Uε is constructed in Chapter 2. We apply it here. It consists of two exponentially decaying boundary layers u0P and u1P and regular polynomial expansion uB , i.e it has a form :

6.2.

359

VARIATIONAL VERSION

u(∞) = uB + u0P + u1P ε

(6.2.16)

where ∞

uB =

εl Nl (

l=0

u0P =

∞ l=0

u1P =

∞ l=0

x2 dl v , ) ε dxl1

x dl v εl Nl0 ( ) l , ε dx1

εl Nl1 (

x1 − 1 x2 dl v , ) l, ε ε dx1

Nl (ξ2 ) are 2 × 2 matrix valued solutions of cell problems ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩

∂Nl ∂ ∂ (A21 Nl−1 ) ∂ξ2 (A22 ∂ξ2 ) + ∂ξ2 ∂ +A12 ∂ξ2 Nl−1 + A11 Nl−2 = hl , ξ2 ∈ (− 12 , 12 ), l + A21 Nl−1 = 0, ξ2 = ± 12 , A22 ∂N 2 ∂ξ1/2 l−l hl = −1/2 (A12 ∂N ∂ξ2 + A11 Nl−2 )dξ2 ,

(6.2.17)

l = 1, 2 . . . ,

Nl0 , Nl1 are some exponentially decaying in the first variable 2×2 matrix valued 1 0 solutions of the chain of the boundary layer problems and N0 = I = , 0 1 hl are diagonal matrices, E 0 h2k+1 = 0, h2 = , 0 1 h4 =

0 0

(λ + 2µ)2 − λ2 0 , , E= −E/12 λ + 2µ

v = (v 1 , v 2 )T is the solution of the homogenized equation 5 E(v 1 )′′ = 0, E (v 2 )′′′′ = 0. − 12

(6.2.18)

Here h11 and h22 are the corresponding elements of the matrix hl . l l j So v and Nl are polynomials and therefore uB has a form

E λ λ ′′′ 3 21 v1 − x2 v2′ + { 16 ( E µ − λ+2µ )x2 + ε 8 ( 3(λ+2µ) − µ )x2 }v2 1 2 ′′ λ λ 2 ′ v2 − λ+2µ x2 v1 + 2(λ+2µ) (x2 − 12 ε )v2

,

(6.2.19)

where v1 (x1 ) = ax1 + b,

(6.2.20)

360

CHAPTER 6. THE MULTI-SCALE DOMAIN DECOMPOSITION v2 (x1 ) = cx31 + dx21 + ex1 + g

(6.2.21)

are polynomials with some undetermined coefficients a, b, c, d, e, g . The constants a, b, c, d, e, g could be defined after calculation of the boundary layers. Introduce the subspace Hε,dec of partially decomposed problem as the subspace of Hε , such that its elements have the form (6.2.19) for all x of the rectangle (δ, 1 − δ) × (− 2ε , 2ε ). The partially decomposed problem has a form: find uε ∈ Hε,dec , such that for all w ∈ Hε,dec , ∂uε ∂w T ) Aij dx + ( ∂ξξj ((0,δ)∪(1−δ,1))×(− ε2 , 2ε ) ∂ξi 1−δ E 2 ′′′ ′′′ E v s )dx1 = + (εEv1′ s′1 + ε3 v2′′ s′′2 + ε5 120µ 2 2 12 δ f wdx, = − ((0,δ)∪(1−δ,1))×(− ε2 , ε2 )

where v and s are polynomials from (6.2.20)-(6.2.21) corresponding to uε and w respectively. Representing the vector valued function uB for x1 ∈ [δ, 1 − δ] as a linear 6 combination i=1 αi Φi , where Φ1 (x1 , x2 ) =

Φ4 (x1 , x2 ) =

Φ6 (x1 , x2 ) =

1 , 0

, Φ2 (x1 , x2 ) =

−x2 , x1

x1

λ )x2 −( λ+2µ

, Φ5 (x1 , x2 ) =

0 , 1

−2x1 x2 λ 2 x1 + ( λ+2µ )(x22 −

,

, Φ3 (x1 , x2 ) =

λ 3 23 − ( λ −3x2 x21 + ( E µ − λ+2µ )x2 + ε 4 3(λ+2µ) 1 λ 2 3 x1 + 3 2(λ+2µ) (x2 − 12 )x1

E µ )x2

ε2 12 )

,

,

,

,

we can deduce the junction conditions on the truncations {x1 = δ} and {x1 = 1 − δ} :

2

x1 =δ,1−δ j=1

ε/2

[A1j

−ε/2

∂udε ]|x Φi (x1 , x2 )dx2 = 0 ∂xj 1

for i = 1, 2, 3, 4, 5, 6. Multiplying the boundary layer functions u0P and u1P by cutting functions 1 ) respectively, we obtain a corrected asymptotic solution χ( x1ε−δ ) and χ( 1−δ−x ε

uaε = uB + u0P χ(

x1 − δ 1 − δ − x1 ) + u1P χ( ), ε ε

6.2.

361

VARIATIONAL VERSION

which satisfies the conditions (i) and (ii) of the subsection 6.2.1 and therefore as in the subsection 6.2.1 we can obtain the estimate (6.2.19), i.e. for any K ˆ ˆ ∈ IR independent of ε that if δ = Kε|ln(ε)| there exist such K then the estimate (6.2.19) holds true. (In order to prove the coercivity (6.2.2) we should use the Korn inequality for thin domains; it gives the estimate (6.2.2) with some r ≤ 3. The main result of this section can be generalized for the rod structures B ε described in the subsection 6.1.3. Consider the equation n

x − yi ∂uε ∂ ), x ∈ Gε , )= fi ( (Arm ε ∂xm ∂xr i=0

(6.2.22)

uε = 0 for x ∈ βˆjε , j = 1, ..., n,

(6.2.23)

where y0 = 0 and yi = 0, i = 1, ..., n, are the ends of the segments ei . We suppose that supp fi belong to the unitary disc {ξ12 + ξ22 ≤ 1}. We state the boundary conditions:

nr Arm

∂uε = 0, for x ∈ ∂B ε \ ∪nj=1 βˆjε . ∂xm

(6.2.24)

Here (n1 , n2 ) is the normal vector. Let Hε be the subspace of vector valued functions of [H 1 (Gε )]2 vanishing on the segments βˆjε , j = 1, ..., n. The variational formulation of the problem (6.2.22)-(6.2.24) is the same as (6.2.15) with Bε instead of Gε . Consider the subspace Hε,dec of Hε , such that its elements have the form

on

e

e

e

0j 0j 3 λ ′′′ 21 −E ( λ v1 − x20j v2′ + { 16 ( E µ )x2 }v2 µ − λ+2µ )(x2 ) + ε 8 3(λ+2µ) e0j 2 e0j ′ 1 2 ′′ λ λ v2 − λ+2µ x2 v1 + 2(λ+2µ) ((x2 ) − 12 ε )v2

ε,δ , B0,j

(6.2.25)

where e

e

v1 (x10j ) = ae0j x10j + be0j , e v2 (x10j )

=

e ce0j (x10j )3

+

e de0j (x10j )2

+

e ee0j x10j

(6.2.26) + ge0j

(6.2.27)

ε,δ is the are polynomials with coefficients ae0j , be0j , ce0j , de0j , ee0j , ge0j . Here B0,j ε,δ part of the cylinders Bj concluded between S0,j and Sj,0 . In this case we can easily check the hypothesis (i) and (ii) and prove the same result as in the case of the rectangle Gε if the hypothesis of section 4.4 are satisfied. This result can be generalized for a multi bundle rod structure (in particular, three dimensional rod structure satisfying to the conditions of Chapter 4). The Korn inequality for rod structures is proved in Appendices 4.A3 and 5.A2. Let us mention that it would be reasonable to apply to the partially decomposed problem the classical domain decomposition method for all 1-dimensional

362


and multi-dimensional parts. This idea allows to parallelize iteratively computations in all these parts. Although in the linear case such parallelization can be obtained by a direct method. The comparison of the direct implementation of the finite elements to problem (6.2.12)-(6.2.14) to the MAPDD finite element implementation is shown in the Figure 6.2.1 (computed by R.Paz and V.Ruas); here the data are taken as follows: ε = 0.05, µ = λ = 1, and f = (0, 1)T . The mesh contains twelve thousand elements.

Figure 6.2.1. The direct and the MAPDD finite element implementation for the elasticity problem in a thin rectangle. Remark 6.2.1 The case when the right hand side f is different from zero and depends on the longitudinal variable in a regular part Gε1 , can be reduced to the previous case. Indeed, consider again the elasticity equation stated in ε ε Gε = (0, 1) × (− , ) : 2 2

∂uε ∂ ) = fε (x1 ), x ∈ Gε , (Arm ∂xm ∂xr

where

(6.2.28)

fε = (f 1 (x1 ), ε2 f 2 (x1 )), f j ∈ C ∞ ([0, 1]).

Let us set the boundary conditions (6.2.13),(6.2.14). The complete asymptotic expansion of the solution Uε again has a form (6.2.16) with the same Nl , Nl0 , Nl1 as above(see (6.2.17)), but v1 , v2 now are the series vα =

∞

j=0

εj vα,j , α = 1, 2,

(6.2.29)

6.2.

363

VARIATIONAL VERSION

asymptotically satisfying the homogenized equations ∞

(l)

εl−2 h11 l v1

= f1 (x1 ),

(6.2.30)

= ε2 f2 (x1 ).

(6.2.31)

l=2

∞

(l)

εl−2 h22 l v2

l=4

Here h11 and h22 are the corresponding elements of the matrix hl from l l (6.2.27). Substituting (6.2.29) into (6.2.30),(6.2.31), we obtain the equations for vα,j : E(v1,j )′′ = −

j−1

(j−k+2) h11 + δj 0 f1 (x1 ) j−k+2 (v1,k )

(6.2.32)

k=0

j−1

−

E (j−k+4) (v2,j )′′′′ = − h22 + δj 0 f2 (x1 ) j−k+4 (v2,k ) 12

(6.2.33)

k=0

It could be proved by induction that (j)

(j)

(v1,j )′′ = Aj f1 , (v2,j )′′′′ = Bj f2 , where A0 = 1/E, B0 = −12/E, EAj = −

j−1

h11 j−k+2 Ak ,

k=0

j−1

−

where j ≥ 1. So,

E h22 Bj = − j−k+4 Bk , 12 k=0

v1′′ =

∞

(j)

εj Aj f1

j=0

v2′′′′ =

∞

(j)

εj Bj f2

(6.2.34)

j=0

and then (m)

v1

=

∞

(j+m−2)

, m≥2

(j+m−4)

, m≥4

εj Aj f1

j=0

(m)

v2

=

∞ j=0

εj Bj f2

(6.2.35)

364


where vα are defined by (6.2.34) up to 6 free constants. Substituting the truncated series (up to εK included) into the variational formulation (6.2.1) we make the changing of unknown functions : we introduce να according to the relations v1 (x1 ) = ν1 (x1 ) +

x1

δ

δ

v2 (x1 ) = ν2 (x1 ) +

x1

δ

δ

θ1

θ2

δ

δ

K θ

(j)

εj Aj f1 (t)dtdθ,

(6.2.36)

j=0

K θ3

(j)

εj Bj f2 (t)dtdθ1 dθ2 dθ3 ,

(6.2.37)

j=0

where the new unknown functions να are polynomials: ν1 (x1 ) = ax1 + b, ν2 (x1 ) = cx31 + dx21 + ex1 + g. So the partially decomposed problem is reduced to the problem of the type (6.2.3) with replacement of vα by the relations (6.2.36), (6.2.37). The same modifications can be done in case of rod structures with the right hand side function depending on the longitudinal variable (as in Chapter 4). The other method to solve the problem with the right hand side function depending on x1 is to consider the standard variational formulation of the problem (6.2.25),(6.2.13),(6.2.14) restricted to the Hilbert subspace Hε,dec of Hε , such that its elements have the form of the truncated series uB , i.e. K+1 l=0

εl Nl (

x2 dl v ) ε dxl1 ,

for all x of the rectangle (δ, 1 − δ) × (− 2ε , 2ε ). Here v is an arbitrary vector valued function of H K+2 ([δ, 1 − δ]), and the matrices Nl are the ”known” (precalculated) solutions of the chain of problems (6.2.27). This approach can be applied for asymptotic partial decomposition of some homogenization problems (see section 6.4). Remark 6.2.2 This method of partial asymptotic decomposition can be applied to multi-structures, [38, 118, 119, 178 ] .

6.3

Decomposition of a flow in a tube structure

Here below we discuss the application of MAPDD for the Navier-Stokes problem (4.5.1)-(4.5.3) with ε standing for µ. For simplicity we consider the case of g = 0. We associate to problem (4.5.1)-(4.5.3) set in B ε with a right hand side of form (4.5.36) the partially decomposed problem. To this end we cut the cylinders at the distance δ = const ε ln (ε) from the nodes by the planes perpendicular to the segments and replace the inner parts of the cylinders by the corresponding

6.3. DECOMPOSITION OF A FLOW IN A TUBE STRUCTURE

365

parts of the segments ej . We obtain the set B ε,δ . Denote Biε,δ the connected truncated part of B ε , containing the node Oi , eij the part of the segment connecting Biε,δ and Bjε,δ ; let Sij be a cross-section of the truncated cylinder corresponding to eij , such that it belongs to ∂Biε,δ .

Figure 6.3.1. A tube structure.

Figure 6.3.2. The asymptotically partially decomposed domain; δ = O(ε|ln(ε)|). Consider the equations for each Biε,δ :

366


ν∆U − (U, ∇)U − ∇P = f,

(6.3.1)

div U = 0, x ∈ Biε,δ

(6.3.2)

U = 0 x ∈ ∂Biε,δ ∩ ∂B ε , i = 1, ..., N,

(6.3.3)

−p′j1 ,j2 = (ΓTj1 ,j2 f )1 − wj1 ,j2

(6.3.4)

U = ε2 wj1 ,j2 Γj1 ,j2 (˜ uej1 ,j2 , 0, ..., 0)T ,

(6.3.5)

the equation for each segment ej1 ,j2 :

and the interface conditions on each cross-section Sj1 ,j2

∂U − (U, n)U − P n), Γj1 ,j2 (˜ uej1 ,j2 , 0, ..., 0)T )ds ∂n Sj1 ,j2 (n, Γj1 ,j2 (˜ uej1 ,j2 , 0, ..., 0)T ) ds pj1 ,j2 = − ((ν

(6.3.6)

Sj1 ,j2

where u ˜ej1 ,j2 is the solution of the Dirichlet problem for Poisson equation (4.5.10) on the cross-section of the rod ej1 ,j2 and Γj1 ,j2 is the matrix of passage to the local base corresponding to the segment ej1 ,j2 with the origin in Oj1 , wj1 ,j2 are e unknown constants, pj1 ,j2 is a function of the variable x1j1 ,j2 defined on each segment ej1 ,j2 . And e

e

pj1 ,j2 (x1j1 ,j2 ) = pj2 ,j1 (|ej1 ,j2 | + 2δ − x1j1 ,j2 ),

(6.3.7)

wj1 ,j2 = − wj2 ,j1 .

(6.3.8)

Taking into account (6.3.5), relation (6.3.6) can be rewritten in a form − ((ε4 wj21 ,j2 u ˜ej1 ,j2 Γj1 ,j2 (˜ uej1 ,j2 , 0, ..., 0)T − Sj1 ,j2

+P n), Γj1 ,j2 (˜ uej1 ,j2 , 0, ..., 0)T )ds = =−

(n, Γj1 ,j2 (˜ uej1 ,j2 , 0, ..., 0)T ) ds pj1 ,j2 .

(6.3.6′ )

Sj1 ,j2

Let us give the variational formulation for this problem. Let ε,δ Hdiv=0 (B1ε,δ , ..., BN , e1 , ..., eM ) be the space of ordered collections (U U1 , ..., UN , w1 , ..., wM ), where Ui are the vector valued functions from Hdiv=0 (Biε,δ ), equal to zero on ∂Biε,δ ∩∂B ε , and w1 , ..., wM are the constants associated with the segments e1 , ..., eM such that for each ej , containing ej1 ,j2 , with Bjε,δ , we have and connecting Bjε,δ 1 2 Uj1 = sign(j2 −j1 )ε2 wj Γj1 ,j2 (˜ uej1 ,j2 , 0, ..., 0)T , on Sj1 ,j2 and on Sj2 ,j1 . (6.3.9)


367

The scalar product is defined as ((U U1 , ..., UN , w1 , ..., wM ), (V V1 , ..., VN , v1 , ..., vM ))P =

M s ∂U Ui ∂Φi ) dx + ε2+(s−1) (|ej |−2δ) (−˜ uej (ξ L ))dξwj vj . , ( ε,δ ∂x ∂x r r βj r=1 j=1 i=1 Bi Note that βj u ˜ej (ξ L )dξ < 0 because u ˜ej (ξ L ) < 0 by the principle of maximum for elliptic equation (4.5.10). The norm is defined as

N

ν

1/2

(U U1 , ..., UN , w1 , ..., wM )P = ((U U1 , ..., UN , w1 , ..., wM ), (U U1 , ..., UN , w1 , ..., wM ))P .

Then the variational formulation is as follows: find (U U1 , ..., UN , w1 , ..., wM ) ∈ ε,δ Hdiv=0 (B1ε,δ , ..., BN , e1 , ..., eM ) and such that for any (Φ1 , ..., ΦN , v1 , ..., vM ) ∈ ε,δ Hdiv=0 (B1ε,δ , ..., BN , e1 , ..., eM ) the integral identity holds true: −

+

N i=1

Biε,δ

N i=1

Biε,δ

ν

s r=1

(

∂U Ui ∂Φi , ) dx + ∂xr ∂xr

(U Ui , (U Ui , ∇)Φi ) dx + ε2+(s−1)

=

N

Biε,δ

i=1

+ ε2+(s−1)

M j=1

βj

u ˜ej (ξ L )dξ

M j=1

(|ej | − 2δ)

u ˜ej (ξ L )dξwj vj

βj

(f, Φ)dx

ej \∪si=1 Biε,δ

e

(ΓTj f )1 dx1j vj .

(6.3.10)

Variational formulation (6.3.10) can be obtained in a following way (cf. secε 0 ε ˆ0 tion 6.2). Consider the subspace H div =0 (B , δ) of the space Hdiv =0 (B ) which 0 ε ˆ ˆ contains all functions U of Hdiv=0 (B , δ) such that for any truncated segment ej1 ,j2 they coincide with the Poiseuille flow uej1 ,j2 , 0, ..., 0)T , on the truncated part of the cylinder forming cj1 ,j2 Γj1 ,j2 (˜ ε B stretched between the cross sections Sj1 ,j2 and Sj2 ,j1 ; cj1 ,j2 is a constant on this truncated cylinder.

368


ε ˆ0 Figure 6.3.3. The structure of space H div =0 (B , δ). Consider the following partially decomposed variational problem: ε ε ˆ ∈H ˆ0 ˆ0 to find U div =0 (B , δ) such that for all Φ ∈ Hdiv =0 (B , δ)

−

Bε

ν

s i=1

(

ˆ ∂Φ ∂U ) dx + , ∂xi ∂xi

Bε

ˆ , (U ˆ , ∇) Φ) dx = (U

(f, ϕ) dx.

Bε

(6.3.11) ε ˆ ∈H ˆ0 (B the ordered collection , δ) Associating to each function U div =0 ε,δ (U U1 , ..., UN , w1 , ..., wM ) ∈ Hdiv=0 (B1ε,δ , ..., BN , e1 , ..., eM ) in such way that ε,δ 2 ˆ ˆ U = Ui on each Bi , and U = ε wj Γj (˜ uej , 0, ..., 0)T on each truncated cylinder forming B ε corresponding to the segment ej (under the convention that each ej stretched between Oj1 and Oj2 , j1 < j2 , has the origin in Oj1 ) we obtain (6.3.10) and respectively (6.3.1)-(6.3.8) as an equivalent formulation of (6.3.11). It can be proved as in [87] that for sufficiently small ε there exists a unique solution to (6.3.11) due to the fixed point theorem and the Poincar´ é - Friedrichs inequality for B ε . The a priori estimate holds true with a constant independent of ε. Suppose that fî in vicinities of nodes are approximated by Taylor’s formula up to the terms O(εK ) as it was done in section 4.5. The above asymptotic analysis of problem (4.5.1)- (4.5.3) (section 4.5) shows that U a satisfies the integral identity (6.3.11) with a discrepancy

Bε

εK (ρ, Φ)dx,

(6.3.12)

369


where ρ is a bounded vector valued function. Moreover, U a does not belong ε ˆ0 to H div =0 (B , δ) because it gives a discrepancy in condition (6.3.5) of order O(exp(−cδ/ε)), c > 0 in C 1 , i.e., U a = ε2 cK uej1 ,j2 , 0, ..., 0)T + exp(−cδ/ε)r0 , ej1 ,j2 Γj1 ,j2 (˜

x ∈ Sj1 ,j2 , (6.3.13)

K l−2 e where r0 is a bounded in C 1 vector valued function, cK cl . Theree = l=2 ε ε ˆ0 fore, we shall slightly change U a in order to obtain a function of H div =0 (B , δ). ε,δ The discrepancy exp(−cδ/ε)r0 can be continued from Sj1 ,j2 to Bj1 in such a way that the divergence of the extension vanishes and that this extension has an order O(exp(−c1 δ/ε)), c1 > 0 in H 1 (Bjε,δ ). Such extension exists (cf. 1 K l−2 ej ε,δ a [87]) because div U = 0 in Bj1 and the constants cK cl ej = l=2 ε ˜ a . Define δU ã = satisfy conditions (4.5.43),(4.5.44). Denote this extension δU U a − ε2 cK uej1 ,j2 , 0, ..., 0)T on truncated parts of cylinders between ej1 ,j2 Γj1 ,j2 (˜ ε ã ∈ H ˜ a = U a − δU ˆ0 the cross sections Sj1 ,j2 and Sj2 ,j1 . Now U div =0 (B , δ), a ˜ δU H 1 (B ε ) = O(exp(−c2 δ/ε)), c2 > 0, and it satisfies integral identity (6.3.11) with discrepancy of order O(exp(−c3 δ/ε)), c3 > 0 , c3 does not depend on small parameters, i.e.,

−

Bε

s ∂ Uã ∂Φ a a ˜ ˜ (f, ϕ) dx + ) dx + (U , (U , ∇) Φ) dx = , ν ( ∂xi ∂xi Bε Bε i=1 +

s

ν

Bε

−

i=1

=

Bε

˜ a ∂Φ ∂ δU , ) dx − ∂xi ∂xi

Bε

Bε

˜ a , ∇) Φ)dx − (U a , (δU

˜ a , (U a − δU ˜ a , ∇) Φ)dx + (δU (f, ϕ) dx +

Bε

+

(

s

εK (ρ, Φ)dx =

Bε

exp(−c3 δ/ε)(ρi ,

B ε i=1

exp(−c3 δ/ε)(ρ0 , Φ)dx +

∂Φ ) dx + ∂xi

εK (ρ, Φ)dx,

(6.3.14)

Bε

where ρi , i = 0, ..., s are vector valued functions bounded in L2 (B ε ) by a constant independent of the small parameters. ˆ independent of ε that if On the other hand, for any K there exist such K ˆ δ = Kε|ln(ε)| then exp(−c2 δ/ε), exp(−c3 δ/ε) = O(εK ). Taking into consideration the discrepancy (6.3.14) and the a priori estimate for (4.5.1)-(4.5.3) we obtain the following estimate: for any K there exist such ˆ independent of ε that if δ = Kε|ln(ε)| ˆ K then

370


ˆ −U ˜ a H 1 (B ε ) = O(εK ). U (6.3.15) K l−2 ej Remark 6.3.1 Defining cK cl we obtain the estimate ej = l=2 ε K K (ua |B ε,δ − U1 , ..., ua |B ε,δ − UN , cK e1 − w1 , ..., ceM − wM )P = O(ε ). 1

N

Combining the estimates (4.5.47) and (6.3.15) we obtain that for any K ˆ independent of ε that if δ = Kε|ln(ε)| ˆ there exists such K then the estimate holds true ˆ − uH 1 (B ε ) = O(εK ). U

(6.3.16)

The estimate (6.3.16) justifies the MAPDD for the Navier-Stokes problem. Remark 6.3.2 The present section is devoted to an approximation of completely three-dimensional (two-dimensional) problem by a hybrid problem that is ”mainly one-dimensional”, i.e., it is of dimension 1 on the major part of the domain. On the other hand another related problem was considered recently by S.A.Nazarov and M.Specovius-Neugebauer [122,184]. They constructed a special approximation of problems in unbounded domain by problems in bounded (truncated) domain. This approach also takes into consideration the information on asymptotic behavior of solution of the problem in unbounded domain at infinity. Remark 6.3.3 Some numerical experiments on MAPDD were developed in case of the Stokes flow in thin domains. The comparison to the direct numerical computation of a solution of the problem shows that for the case when the viscosity ν is finite, the boundary layers are very narrowly localized in the neighborhoods of the junctions, so δ can be taken almost equal to just ε. However, if the Reynolds number becomes greater (for example if the viscosity is small) then the multi-dimensional parts become also greater. The same remark holds for the flows in wavy tubes and for the extrusion process modelling. The Stokes flow in an extruder with a screw of a form presented in Figure 6.3.4 was computed by I.Sirakov using the asymptotic domain decomposition on five subdomains, as it is shown in the Figure 6.3.5. The pressure and the particles trajectories were calculated (Figure 6.3.6).


371

Figure 6.3.4. The form of the screw.

Figure 6.3.5. The asymptotic partial domain decomposition for the screw.

372


Figure 6.3.6. A typical trajectory of particles.

Figure 6.3.7. Extrusion process.

373

6.4. THE PARTIAL HOMOGENIZATION

6.4

The partial homogenization

This section is devoted to an application of MAPDD to the homogenization problem. The idea of partial homogenization on some subdomain was discussed in [156]: we replace the initial equation by the homogenized equation of high order on the main part of the domain and we keep the initial equation in some boundary strip. The main question is how to conjugate these two equations (probably of different order) to obtain the accuracy of given power of ε.

Figure 6.4.1. The partial homogenization. Here below we consider the model problem in a layer with Dirichlet boundary condition, and we propose and justify a method of ”conjugation” of the higher order homogenized equation with the initial equation. Consider the homogenization of a boundary value problem in a layer [132],[133]: s

k,j=1

x′ ∂uε ∂ ) = f (x1 ), x1 ∈ (0, d), (Akj ( ) ε ∂xj ∂xk

uε |x1 =0 = 0,

uε |x1 =d = 0,

(6.4.1) where x = (x1 , x′ ), x′ = (x2 , . . . , xs ), Akj (ξ ′ ) are 1 - periodic in ξ ′ ∈ IRs−1 functions, satisfying the following conditions:

(i)∃κ0 > 0, ∀ξ ′ ∈ IRs−1 , ∀η ∈ IRs , η = (η1 , . . . , ηs ),

s

k,j=1

Akj (ξ ′ )ηj ηk ≥ κ0

(ii) ∀ξ ′ ∈ IRs−1 , k, j ∈ {1, . . . , s}, Akj (ξ ′ ) = Ajk (ξ ′ ),

s i=1

ηi2

374


f is a C ∞ - regular function. Let T be a finite number, multiple of ε. Assume that Akj are piecewise smooth functions in the sense [16]. Then there exists a unique T −periodic in x2 , . . . , xs , solution to this problem (see [16]); T is a multiple of ε. This solution is a solution of the following variational formulation. Let H01,per′ be a space that is completion of the space of C ∞ − regular T −periodic in x2 , . . . , xs functions vanishing if x1 = 0 or x1 = d with respect to the norm H 1 ((0, d) × (0, T )s−1 ); then uε is sought as function of H01,per′ satisfying

J(u, ϕ) =

Ωd,T

*

s

k,j=1

6 x′ ∂u ∂ϕ + f ϕ dx = 0, ∀ϕ ∈ H01,per′ , (6.4.2) Akj ( ) ε ∂xj ∂xk

where Ωd,T = (0, d) × (0, T )s−1 . The asymptotic expansion of the solution to problem (6.4.1) was constructed in [133]. It has a form uεa

(K)

x (x1 − d, x′ ) (K) (K) )+ = uBLO (x, ) + uBLd (x, ε ε

+

K+1

εl N l (

l=0

x′ l (K) )D1 vε (x1 ), ε

vε(K) (x1 ) =

K+1

εj vj (x1 ),

(6.4.3)

(6.4.4)

j=0

(K)

(K)

where the ”boundary layers” uBLO and uBLd are exponentially decaying func(K) (K) tions such that for all x ∈ [0, d]×IRs−1 , |uBLO (x, ξ)|, |uBLd (x, ξ)| ≤ C1 e−C2 |ξ1 | , ′ C1 , C2 > 0, C1 , C2 do not depend on ε; Nl (ξ ) are 1-periodic functions, solutions of the sequence of cell problems: Lξξ N1 = −

Lξξ Nl = −

−

s

A1k

k=2

+
, ∂ξk

(6.4.5)

375


here =

(0,1)s−1

0, N0 = 1, D1 =

′

dξ , Lξξ = ∂ ∂x1 .

l ≥ 2,

s

(6.4.6)

∂ ′ ∂ k,j=2 ∂ξk (Akj (ξ ) ∂ξj );

< Nl >= 0 for l >

1 These cell problems are set in variational formulation in Hper , that is the ∞ completion of the space of C −regular 1−periodic functions with respect to the norm H 1 ((0, 1)s ). This sequence of problems is solved recurrently for l = 1, 2, . . . , K + 1.

Functions vj are defined in another sequence of problems (see [133], [16] ) but we do not need this information for the further developments. Let us describe the method of asymptotic partial decomposition of domain for problem (6.4.1) (in formulation (6.4.2)). Consider the subspace Hε,δ, dec of the space H01,per′ that consists of all functions u of H01,per′ having for all x1 ∈ [δ, d − δ] the following presentation in the form of Bakhvalov’s ansatz [16]:

u(x) =

K+1

εl N l (

l=0

x′ l )D1 v, ε

(6.4.7)

where v ∈ H K+2 ((δ, d − δ)). So every u ∈ Hε,δ,dec is related by (6.4.7) to some v ∈ H K+2 ((δ, d − δ)), so that we can consider a couple (u, v). ˆ | ln ε |, where K ˆ is some positive number independent of ε. Let δ = Kε Consider the partially decomposed variational problem J(u, ϕ) = 0,

K+2 . ∀ϕ ∈ Hε,δ,dec

(6.4.8)

If ud is its solution then estimate (6.2.9) proves that for any K ∈ (0, ∞), there ˆ such that if δ = K ˆ ε | ln ε | then the estimate holds exist K uε − ud H 1 ((0,d)×(0,T )s−1 ) = O(εK+1 ).

(6.4.9)

In the layer (δ, d − δ) × IRs−1 ud is presented by formula (6.4.7) as well as ϕ=

K+1 l=0

εl Nl (

x′ l )D1 w, ε

w ∈ HTK+2 per ((δ, d − δ))

(6.4.10)

In all previous examples the subspace Hε,δ,dec was closed in the Hε , while now it is not generally the case. So, the existence of ud should be studied separately.

376


Let us prove now the existence and the uniqueness of the solution ud of problem (6.4.8). To this end let us calculate integrals

I1 (u, ϕ) =

s

{

Akj (

(δ,d−δ)×(0,T )s−1 k,j=1

and

I2 (ϕ) =

x′ ∂u ∂ϕ }dx . ) ε ∂xj ∂xk

f (x1 )ϕdx

(δ,d−δ)×(0,T )s−1

K+2 for u, ϕ ∈ Hε,δ,dec , satisfying (6.4.7) and (6.4.10). We get

I1 (u, ϕ) = T s−1

K+2

(δ,d−δ) l,m=1

I2 (ϕ) = T s−1

˜ l,m Dl vDm wdx1 , εl+m−2 h 1 1 f (x1 )w(x1 )dx1 ,

(δ,d−δ)

where ˜ l,m = h

s

< Akj (ξ ′ )(

k,j=1

∂ ∂ Nm + δk1 Nm−1 ) > Nl + δj 1 Nl−1 )( ∂ξk ∂ξξj

defined by the solutions of cell problems (6.4.5),(6.4.6). The terms containing derivatives ∂ξ∂k are dropped if k = K +2, and ∂ξ∂ j are dropped if l = K +2; these coefficients were introduced in [183]. This presentation holds true due to the separation of variables x′ and x1 in (6.4.7) and (6.4.10) and because < Nl >= 0 for all l > 0. Below ε is a fixed positive number small enough, such that the ratio T /ε is integer. K+2 Consider space Hε,δ,dec × H K+2 ((δ, d − δ)) supplied by the following inner product for K+2 (u, v), (ϕ, w) ∈ Hε,δ,dec × H K+2 ((δ, d − δ))

related by (6.4.7) and (6.4.10): a(u, v; ϕ, w) =

{

s

Akj (

GBL k,j=1

+T

s−1

K+2

(δ,d−δ) l,m=1

=

{

s

x′ ∂u ∂ϕ }dx + . ) ε ∂xj ∂xk

˜ l,m Dl vDm wdx1 = εl+m−2 h 1 1

Ωd,T k,j=1

Akj (

x′ ∂u ∂ϕ }dx . ) ε ∂xj ∂xk

where GBL = (0, δ) × (0, T )s−1 ∪ (d − δ, d) × (0, T )s−1 .

(6.4.11)

377


Let us check that it is really an inner product. Indeed, its bi-linearity and symmetry properties are evident; its non-negativeness is a consequence of the non-negativeness of the quadratic part of J(u, u). Assume now that a(u, this quadratic part of J(u, u) vanishes and so u = 0, then v; u, v) = 0.′ Then ′ u(x , x )dx = 0 for x1 ∈ (δ, d − δ); therefore (as < Nl >= 0 for l > 0) 1 (0,T )s−1 v(x1 ) = 0 for x1 ∈ (δ, d − δ). Problem (6.4.8) is equivalent to the following problem: find a couple K+2 (ud , vd ), ∈ Hε,δ,dec × H K+2 ((δ, d − δ))

related by (6.4.7)such that for any couple K+2 (ϕ, w) ∈ Hε,δ,dec × H K+2 ((δ, d − δ))

related by (6.4.10),

a(u, v; ϕ, w) +

f (x1 )ϕdx + T s−1

f (x1 )w(x1 )dx1 = 0.

(δ,d−δ)

GBL

Let us check the continuity of the linear functional f (x1 )ϕdx − T s−1 f (x1 )w(x1 )dx1 . Φ(ϕ, w) = − (δ,d−δ)

GBL

We have |Φ(ϕ, w)| ≤ T

s−1 2

f L2 (0,d) {ϕL2 (GBL ) + T

s−1 2

wL2 (δ,d−δ) }.

Here δ ϕL2 (GBL ) ≤ √ ∇ϕL2 (GBL ) ≤ 2 s x′ ∂ϕ ∂ϕ δ }dx ≤ . { Akj ( ) ≤ √ ε ∂xj ∂xk 2κ0 GBL k,j=1

≤ √

T

= T

s−1 2

− s−1 2

wL2 (δ,d−δ) = T

,

δ

d−δ

(

(0,T )s−1

δ a(ϕ, w; ϕ, w). 2κ0

− s−1 2

,

d−δ

(0,T )s−1

δ

ϕ(x)dx′ )2 dx

(

1

≤

,

w(x1 )dx′ )2 dx1 =

(δ,d−δ)×(0,T )s−1

(ϕ(x))2 dx =

378


= ϕL2 ((δ,d−δ)×(0,T )s−1 ) ≤ √

d a(ϕ, w; ϕ, w), 2κ0

by the Poincar´ é Friedrichs inequality for ϕ. So Φ is a continuous functional. K+2 × H K+2 ((δ, d − δ)) Consider problem (6.4.8) set on the completion of Hε,δ,dec with respect to the norm a(•, •; •, •). Applying now the Riesz theorem we get the existence and the uniqueness of solution. Remark 6.4.1 Let Akj be constant coefficients equal to δkj . Then Nl = 0 ˜ lm = δl1 δm1 and therefore the completion of for all l > 0 and N0 = 1, and so h K+2 K+2 Hε,δ,dec × H ((δ, d − δ)) is a subspace of space H01 per′ × H 1 ((δ, d − δ)) such ∂u =0 that, for all x1 ∈ (δ, d − δ), u(x) = v(x1 ), where v ∈ H 1 ((δ, d − δ)), ie., ∂x j for all x1 ∈ (δ, d − δ), j = 2, ..., s. 2 2 ˜ Assume now that < NK +2 >= 0. Then hK+2,K+2 =< A11 NK +2 > is positive. Varying w ∈ H K+2 ((δ, d − δ)) we prove that v satisfies on (δ, d − δ) the ordinary differential equation K+2

l,m=1

˜ l,m Dl+m v(x1 ) + f (x1 ) = 0, x1 ∈ (δ, d − δ). εl+m−2 (−1)m h 1

(6.4.12)

This v belongs to C ∞ ([δ, d − δ]) because f ∈ C ∞ ([0, d]) and therefore ud ∈

K+2 Hε,δ,dec .

Applying now estimate (6.2.9) we get the following theorem. Theorem 6.4.1. The estimate holds true uε − ud H 1 ((0,d)×(0,T )s−1 ) = O(εK+1 ).

(6.4.13)

The solution of problem (6.4.8) satisfies equation (6.4.1) for x ∈ GBL and boundary conditions (6.4.1) for x1 = 0 and for x1 = d, it satisfies as well equation (6.4.12) for vd and the 2K + 4 interface conditions for x1 = δ and for x1 = d − δ. Let us give these conditions for x1 = δ; for x1 = d−δ they are the same. Firstly, ud and v satisfy conditions (6.4.7) for x1 = δ (and for x1 = d − δ). And then there are K + 2 interface conditions that are the consequences of the integration by parts and of the equating the coefficients of D1r w(δ), r = 0, ..., K + 1 : T 1−s

s

(0,T )s−1 j=1

=

K+2 K+2

Akj (

x′ x′ ∂ud |x1 =δ Nr ( )dx′ = ) ε ε ∂xj

˜ l,m Dl+m−r−1 vd (δ). εl+m−2 (−1)m−r−1 h 1

(6.4.14)

l=1 m=r+1

Remark 6.4.2 Another way to prove the existence of solution of (6.4.8) is to prove that ”the classical solution” of problem (2.1) in GBL , satisfying

379


boundary conditions (6.4.1) for x1 = 0 and for x1 = d, equation (6.4.12) for vd and interface conditions (6.4.7) and (6.4.14) for x1 = δ and for x1 = d − δ is a solution to (6.4.8). In order to find this classical solution let us find first a T −periodic in x′ solution ud0 of equation (6.4.1) in GBL and vanishing for x1 = 0, δ, d − δ and d, and a solution vd0 of equation (6.4.12) satisfying conditions D1l v(δ) = 0, D1l v(d − δ) = 0, l = 0, ..., K + 1. Then we seek vd in a form vd = vd0 +

2K+4

cr Vr ,

(6.4.15)

r=1

where {V Vr , r = 1, ..., 2K + 4} is a basis of the space of solutions of the homogeneous ordinary differential equation K+2

l,m=1

˜ l,m Dl+m v(x1 ) = 0, x1 ∈ (δ, d − δ). εl+m−2 (−1)m h 1

(6.4.16)

and ud = ud0 +

2K+4

cr udr ,

(6.4.17)

r=1

where udr are the T −periodic in x′ solutions to problems s

k,j=1

x′ ∂udr ∂ ) = 0, x ∈ GBL , (Akj ( ) ε ∂xj ∂xk

udr =

K+1

εl N l (

l=0

udr |x1 =0 = 0,

udr |x1 =d = 0,

x′ l )D1 Vr , x1 = δ, d − δ. ε

(6.4.18)

So now problem is reduced to an algebraic linear system of equations for c1 , ..., c2K+4 , such that ud and vd satisfy (6.4.15), (6.4.17) and (6.4.14) for x1 = δ and for x1 = d − δ. Its matrix is positive definite because the corresponding quadratic form

a(ud0 +

2K+4 r=1

cr udr , vd0 +

2K+4 r=1

cr Vr ; ud0 +

2K+4 r=1

cr udr , vd0 +

2K+4

cr Vr )

r=1

is positive definite and Vr are linearly independent functions on (δ, d − δ). Consider an example K = 0. We have then

380


a(u, v; ϕ, w) =

s

Akj (

GBL k,j=1

+

2

˜ lm εl+m−2 h

GI l,m=1

x′ ∂u ∂ϕ dx+ ) ε ∂xj ∂xk

∂lv ∂mw dx, ∂xl1 ∂xm 1

(6.4.19)

where ˜ 11 = Aˆ11 = h

s

< A1j

j=1

˜ 12 = h ˜ 21 = h

s

< A1j

j=1

∂(N N1 + ξ1 ) >; ∂ξξj

∂(N N1 + ξ1 ) N1 >; ∂ξξj

˜ 22 =< A11 N 2 > . h 1 Integrating identity (6.4.8) by parts and taking into account that w is an arbitrary function of the space H 2 ((δ, d − δ)), we see that v is a solution to the ˜˜ equation of the fourth order (if h 22 = 0 ) ˜ 22 ε2 h

2 ∂4v ˜ 11 ∂ v = −f (x1 ), −h 4 ∂x21 ∂x1

x1 ∈ (δ, d − δ)

(6.4.20)

with the interface conditions following from the equations

{x1 =δ,x2 ,...,xs ∈[0,T ]}

˜ 11 {h

3 2 ∂v ˜ 22 ∂ v w(δ)}dx′ = ˜ 12 ∂v ) ∂w (δ)−ε2 h ˜ 22 ∂ v +εh w(δ)+(ε2 h ∂x31 ∂x1 ∂x1 ∂x21 ∂x1

s

{x1 =δ,x2 ,...,xs ∈[0,T ]} j=1

and

Aij

x ∂w ∂u ¯d (δ))dx′ (w(δ) + εN N1 ( ) ε ∂x1 ∂xj

x ∂v u ¯d (δ, x2 , . . . , xs ) = v(δ) + εN1 ( ) (δ); ε ∂x1 i.e. for x1 = δ denoting Sδ = {x1 = δ, x2 , . . . , xs ∈ [0, T ]}, we get ⎧ s ∂u ¯d 1 ′ 2 ˜ ∂3v ˜ ∂v ⎪ ⎨ h11 ∂x1 − ε h22 ∂x31 = T s−1 Sδ j=1 A1j ∂xj dx , 2 s ∂u ¯d x 1 ′ ˜ 22 ∂ v2 + εh ˜ 12 ∂v = s−1 ε2 h j=1 A1j ∂xj N1 ( ε )dx , ∂x1 T Sδ ∂x1 ⎪ ⎩ ∂v (δ). N1 ( δε , xε2 , . . . ; xεs ) ∂x u ¯d (δ, x2 , . . . , xs ) = v(δ) + εN 1

(6.4.21)

The same interface conditions we have on the surface Sd−δ = {x1 = d − δ, x2 , . . . , xs ∈ [0, T ]}; δ in these conditions has to be replaced by d − δ. In GBL we keep equation (6.4.1) for u ¯d . So, (6.4.1) in GBL for u ¯d , (6.4.20) in GI for v,

381


and interface conditions (6.4.21) on Sδ and on Sd−δ constitute the differential version of partially homogenized problem (6.4.8) for K = 0. Estimates (6.4.13) give the error of order O(ε). This estimate improves the standard estimate for the difference of the exact solution and of the first order √ asymptotic solution without the boundary layer corrector that is of order O( ε). The idea of the partial homogenization can be applied to the creation of hybrid semi-discrete (semi-continuum) models. Such models are important for nano-structures. The nano-structures are often constituted of a very thin layer (coating) on a homogeneous body. The thickness of the coating is of order of some hundreds of atomic sizes, and the properties change very rapidly across the coating. Therefore it is reasonable to describe the field in the coating layer by a discrete model, similar to some difference scheme, while the main material of course should be described by some continuum model, that is, for example, a partial differential equation. The interface conditions between these two models could be obtained using the ideas of the partial asymptotic domain decomposition. Let us consider a very simple example of a model problem of this kind. Consider the finite difference scheme [173]

(aε yx )x¯ = fi ,

i = 1, ..., N − 1,

y0 = 0, yN = 0.

(6.4.22)

Here N is an integer number, y = (y0 , ..., yN ) stands for an unknown vector with the components describing the field in the nodes of the atomic grill xi = ih, where h = N1 is the small parameter, the step of the grill, ie. the distance between atoms; fi are given real numbers, the right hand side; ε is a small parameter, characterizing the variation scale of the coefficient in the coating layer, ε ≥ h. We use here the notations from [173]: for any vector w = (w0 , ..., wN ), we define the vector wx = (wx 0 , ..., wx N −1 ), such that, wx i = wi+1h−wi ; in the same way for any vector w = (w0 , ..., wN ), we define i−1 the vector wx¯ = (wx¯ 1 , ..., wx¯ N ), such that, wx¯ i = wi −w ; aε is a given coh efficient Kε (x) that is multiplied by yx according the following rule: aε yx is an N −dimensional vector with the components Kε (xi + 0.5h)yx i , i = 0, .., N − 1. Assume that for some given finite M, Kε (x) is defined as K( xε ) if x ≤ M ε, and as 1 if x ≥ M ε, where K is a continuous positive function, such that, there exists a positive κ, such that, K(ξ) ≥ κ, K(M ) = 1. So, equation (6.4.22) has a form

Kε (xi + 0.5h)(yi+1 − yi ) − Kε (xi − 0.5h)(yi − yi−1 ) = h2 = fi ,

i = 1, ..., N − 1,

y0 = 0, yN = 0.

(6.4.23)

Assume that fi = f (xi ), f ∈ C 2 [M ε, 1] and it is a restriction to [M ε, 1] of a function independent of ε and h. It is impossible to pass to the limit everywhere in (6.4.23) as ε → 0, because the coefficients and the right hand side vary rapidly in [0, M ε]. Let us consider the partially homogenized problem for an unknown function u defined in [Lh, 1] and an unknown vector y d ∈ IRL , for an integer L, that is,

382


(aε yxd )x¯ = fi ,

i = 1, ..., L − 1,

L − 1 ≥ (M + 1)ε/h,

u”(x) = f (x), x ≥ Lh, y0d = 0,

u(1) = 0, d , u(Lh) = yL

(u′ + 12 hu”)(Lh) − h

d d yL −yL−1 h

= fL .

(6.4.24)

ie. u′ (Lh) − h

d d yL −yL−1 h

=

1 fL . 2

Consider the function yid extended to the grill ωh = {xi = ih, i = 0, ..., N } as follows: yid = u(xi ) if x ≥ Lh. Then the extended yid satisfies the relations (aε yxd )x¯ = fi + ri ,

d i = 1, ..., N − 1, y0d = 0, yN = 0, √ N −1 where ri = 0 for i = 1, ..., L − 1, and r = h i=L ri2 = O( h h ), and so, [173], √ y − y d (1) = O( ),

where y(1)

N −1 (y 2 + (y = h i

2 x i ) ).

i=0

This estimate justifies the closeness of discrete model (6.4.23) and hybrid multi-scale model (6.4.24).

6.5


A great number of applied problems contain small parameters. Normally their presence either in the equation or in the domain makes the numerical implementation more complicated, more time and memory consuming. This issue emphasizes the importance of the asymptotic methods studying the behavior of the solution as the small parameter tends to zero. Nevertheless the asymptotic methods are often related to some cumbersome calculations, or they are not too comprehensible for non-specialists. That is why some special numerical methods taking into account the asymptotic behavior of the solution were developed.


383

One of such ideas has been implemented in the numerical schemes uniform with respect to the small parameter [11],[64],[181]; in the case of multi-scale problems the idea of the super-elements, hierarchic modelling or the two-scale finite element methods is developed [185],[51],[10],[180]; another approach is to prescribe some special modified boundary conditions (the so called artificial boundary conditions) in order to increase the accuracy of the approximate solution [122],[123]. The method of asymptotic partial decomposition of domain also belongs to a class of methods taking into account the structure of the asymptotic solution of the problem. It was proposed by the author in [154],[156] (the differential version), [160],[161] (the variational version). It was applied to the conductivity and the elasticity equations [161], to the Stokes [28] and the NavierStokes equation [157],[158],[162]. The finite element implementation is studied in [56],[57] and [28]. The comparison of such an implementation to the adaptive mesh methods shows that the MAPDD is not worse than the adaptive meshes because the last needs some iterations to create the final mesh. The variational version of MAPDD generalizes the idea of projection of the variational formulation on a subspace of functions having the structure of the regular asymptotic expansion. MAPDD differs from this method by keeping the initial formulation in the boundary layer zone, and so, by the idea of special asymptotic decomposition of domain. This difference is especially important, for example, in the case of thin domains of a complex structure, such as the finite rod structures.

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397 Subject index

asymptotic partial domain decomposition 337 solution (formal) 23 boundary layer corrector 26 boundary layer corrector in homogenization 32 techniques 26 cell problem 5,39,120 code EFMODUL 6,120 conductivity equation 2,37 contrasting coefficients 110 effective coefficients 5,6 elasticity equation 2,57 equivalent homogeneous plate 154 FL-convergence 168, 251 formal asymptotic solution (f.a.s.) 23 finite element method 346 finite rod structures 12,162 heterogeneous plate 130 rod 21 homogenization method 9, 22 homogenized boundary conditions 26,33 equation 25 of high order 25,31 high order homogenization method 22 Korn inequality 18, 243, 321 lattice (skeletal) structure 13, 248 L-convergence 15, 167, 250 macroscopic (slow) variable 9 method of asymptotic partial decomposition of domain 337 microscopic (fast) variable 9 multi-component homogenization 110 multi-scale models ix, 337 Navier-Stokes equation 217 net 270 non-stationary (time dependent, non-steady state) model 98 elasticity equation 103 partial homogenization 373 plate 130 Poincar´ é inequality 17, 233, 318 Poincar´ é - Friedrichs inequality 17, 233, 251, 318 random coefficients 307 rectangular lattice 13, 249 rod (bar) 21

SUBJECT INDEX

398 section 161, 248 shape optimization 173 steady-state (stationary) model 36,57 stress analysis 9 tube (pipe) structure 215

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