Numerical methods for elliptic partial differential equations Arnold Reusken

Transcription

1 Numerica methods for eiptic partia differentia equations Arnod Reusken

2 Preface This is a book on the numerica approximation of partia differentia equations. On the next page we give an overview of the structure of this book:

3 Eiptic boundary vaue probems (chapter 1): Poisson equation: scaar, symmetric, eiptic. Convection-diffusion equation: scaar, nonsymmetric, singuary perturbed. Stokes equation: system, symmetric, indefinite. Weak formuation (chapter ) Finite eement method Basic principes (chapter 3); appication to Poisson equation. Streamine-diffusion FEM (chapter 4); appication to convection-diffusion eqn. FEM for Stokes equation (chapter 5). Iterative methods Basics on inear iterative methods (chapter 6). Preconditioned CG method (chapter 7); appication to Poisson equation. Kryov subspace methods (chapter 8); appication to convection-diffusion eqn. Mutigrid methods (chapter 9). Iterative methods for sadde-point probems (chapter 10); appication to Stokes equation. Adaptivity A posteriori error estimation (chapter ). Grid refinement techniques (chapter ). 3

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5 Contents 1 Introduction to eiptic boundary vaue probems Preiminaries on function spaces and domains Scaar eiptic boundary vaue probems Formuation of the probem Exampes Existence, uniqueness, reguarity The Stokes equations Weak formuation 17.1 Introduction Soboev spaces The spaces W m (Ω) based on weak derivatives The spaces H m (Ω) based on competion Properties of Soboev spaces Genera resuts on variationa formuations Minimization of functionas and sadde-point probems Variationa formuation of scaar eiptic probems Introduction Eiptic BVP with homogeneous Dirichet boundary conditions Other boundary conditions Reguarity resuts Riesz-Schauder theory Weak formuation of the Stokes probem Proof of the inf-sup property Reguarity of the Stokes probem Other boundary conditions Gaerkin discretization and finite eement method Gaerkin discretization Exampes of finite eement spaces Simpicia finite eements Rectanguar finite eements Approximation properties of finite eement spaces Finite eement discretization of scaar eiptic probems Error bounds in the norm Error bounds in the norm L Stiffness matrix Mass matrix

6 3.6 Isoparametric finite eements Nonconforming finite eements Finite eement discretization of a convection-diffusion probem Introduction A variant of the Cea-emma A one-dimensiona hyperboic probem and its finite eement discretization The convection-diffusion probem reconsidered We-posedness of the continuous probem Finite eement discretization Stiffness matrix for the convection-diffusion probem Finite eement discretization of the Stokes probem Gaerkin discretization of sadde-point probems Finite eement discretization of the Stokes probem Error bounds Other finite eement spaces Linear iterative methods Introduction Basic inear iterative methods Convergence anaysis in the symmetric positive definite case Rate of convergence of the SOR method Convergence anaysis for reguar matrix spittings Perron theory for positive matrices Reguar matrix spittings Appication to scaar eiptic probems Preconditioned Conjugate Gradient method Introduction Conjugate Gradient method Introduction to preconditioning Preconditioning based on a inear iterative method Preconditioning based on incompete LU factorizations LU factorization Incompete LU factorization Modified incompete Choesky method Probem based preconditioning Preconditioned Conjugate Gradient Method Kryov Subspace Methods Introduction The Conjugate Gradient method reconsidered MINRES method GMRES type of methods Bi-CG type of methods

7 9 Mutigrid methods Introduction Mutigrid for a one-dimensiona mode probem Mutigrid for scaar eiptic probems Convergence anaysis Introduction Approximation property Smoothing property Mutigrid contraction number Convergence anaysis for symmetric positive definite probems Mutigrid for convection-dominated probems Nested Iteration Numerica experiments Agebraic mutigrid methods Noninear mutigrid Iterative methods for sadde-point probems Bock diagona preconditioning Appication to the Stokes probem A Functiona Anaysis 35 A.1 Different types of spaces A. Theorems from functiona anaysis B Linear Agebra 41 B.1 Notions from inear agebra B. Theorems from inear agebra

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9 Chapter 1 Introduction to eiptic boundary vaue probems In this chapter we introduce the cassica formuation of scaar eiptic probems and of the Stokes equations. Some resuts known from the iterature on existence and uniqueness of a cassica soution wi be presented. Furthermore, we briefy discuss the issue of reguarity. 1.1 Preiminaries on function spaces and domains The boundary vaue probems that we consider in this book wi be posed on domains Ω R n, n = 1,,3. In the remainder we aways assume that Ω is open, bounded and connected. Moreover, the boundary of Ω shoud satisfy certain smoothnes conditions that wi be introduced in this section. For this we need so-caed Höder spaces. By C k (Ω), k N, we denote the space of functions f : Ω R for which a (partia) derivatives D ν f := ν f x ν, ν = (ν 1 1,...,ν n ), ν = ν ν n, 1... xνn n of order ν k are continuous functions on Ω. The space C k ( Ω), k N, consists of a functions in C k (Ω) C( Ω) for which a derivatives of order k have continuous extensions to Ω. Since Ω is compact, the functiona f max max ν k x Ω D ν f(x) = max ν k Dν f, Ω =: f C k ( Ω) defines a norm on C k ( Ω). The space (C k ( Ω), C k ( Ω) ) is a Banach space (cf. Appendix A.1). Note that f max ν k D ν f,ω does not define a norm on C k (Ω). For f : Ω R we define its support by supp(f) := {x Ω f(x) 0 }. The space C0 k(ω), k N, consists of a functions in Ck (Ω) which have a compact support in Ω, i.e., supp(f) Ω. The functiona f max ν k D ν f,ω defines a norm on C0 k (Ω), but 9

10 (C k 0 (Ω), C k (Ω)) is not a Banach space. For a compact set D R n and λ (0,1] we introduce the quantity [f] λ,d := sup{ f(x) f(y) x y λ x,y D, x y } for f : D R. We write f C 0,λ ( Ω) and say that f is Höder continuous in Ω with exponent λ if [f] λ, Ω <. A norm on the space C 0,λ ( Ω) is defined by f f C( Ω) + [f] λ, Ω. We write f C 0,λ (Ω) and say that f is Höder continuous in Ω with exponent λ if for arbitrary compact subsets D Ω the property [f] λ,d < hods. An important specia case is λ = 1: the space C 0,1 ( Ω) [or C 0,1 (Ω)] consists of a Lipschitz continuous functions on Ω [Ω]. The space C k,λ ( Ω) [C k,λ (Ω)], k N, λ (0,1], consists of those functions in C k ( Ω) [C k (Ω)] for which a derivatives D ν f of order ν = k are eements of C 0,λ ( Ω) [C 0,λ (Ω)]. On C k ( Ω) we define a norm by f f C k ( Ω) + [D α f]. λ, Ω Note that α =k C k,λ ( Ω) C k ( Ω) for a k N, λ (0,1], C k,λ ( Ω) C k,λ 1 ( Ω) for a k N, 0 < λ 1 λ 1, and simiary with Ω repaced by Ω. We use the notation C k,0 ( Ω) := C k ( Ω) [C k,0 (Ω) := C k (Ω)]. Remark The incusion C k+1 ( Ω) C k,λ ( Ω), λ (0,1], is in genera not true. Consider n = and Ω = {(x,y) 1 < x < 1, 1 < y < x }. The function { (sign x)y 11 if y > 0, f(x,y) = 0 otherwise, beongs to C 1 ( Ω), but f / C 0,λ ( Ω) if λ ( 3 4,1]. Based on these Höder spaces we can now characterize smoothness of the boundary Ω. Definition 1.1. For k N, λ [0,1] the property Ω C k,λ (the boundary is of cass C k,λ ) hods if at each point x 0 Ω there is a ba B = {x R n x x 0 < δ, δ > 0 } and a bijection ψ : B E R n such that ψ(b Ω) R n + := {x Rn x n > 0 }, ψ(b Ω) R n +, ψ C k,λ (B), ψ 1 C k,λ (E). (1.1a) (1.1b) (1.1c) For the case n = this is iustrated in Figure??. Figure 1 10

11 A very important specia case is Ω C 0,1. In this case a transformations ψ (and their inverses) must be Lipschitz continuous functions and we then ca Ω a Lipschitz domain. This hods, for exampe, if Ω consists of different patches which are graphs of smooth functions (e.g., poynomias) and at the interface between different patches interior anges are bounded away from zero. In Figure?? we give an iustration for the two dimensiona case. Figure A domain Ω is convex if for arbitrary x,y Ω the incusion {tx+(1 t)y t [0,1] } Ω hods. In amost a theoretica anayses presented in this book it suffices to have Ω C 0,1. Moreover, the domains used in practice usuay satisfy this condition. Therefore, in the remainder of this book we aways consider such domains, uness stated otherwise expicity. Assumption In this book we assume that the domain Ω R n is such that Ω is open, connected and bounded, Ω is of cass C 0,1. One can show that if this assumption hods then C k+1 ( Ω) C k,1 ( Ω) (cf. remark 1.1.1). 1. Scaar eiptic boundary vaue probems 1..1 Formuation of the probem On C (Ω) we define a inear second order differentia operator L as foows: Lu = n i,j=1 a ij u + x i x j n i=1 b i u x i + cu, (1.) with a ij, b i and c given functions on Ω. Because u x i x j = u x j x i we may assume, without oss of generaity, that a ij (x) = a ji (x) hods for a x Ω. Corresponding to the differentia operator L we can define a partia differentia equation Lu = f, (1.3) with f a given function on Ω. In (1.) the part containing the second derivatives ony, i.e. n i,j=1 a ij u, x i x j is caed the principa part of L. Reated to this principa part we have the n n symmetric matrix A(x) = (a ij (x)) 1 i,j n. (1.4) 11

12 Note that due to the symmetry of A the eigenvaues are rea. These eigenvaues, which may depend on x Ω, are denoted by λ 1 (x) λ (x)... λ n (x). Hyperboicity, paraboicity, or eipticity of the differentia operator L is determined by these eigenvaues. The operator L, or the partia differentia equation in (1.), is caed eiptic at the point x Ω if a eigenvaues of A(x) have the same sign. The operator L and the corresponding differentia equation are caed eiptic if L is eiptic at every x Ω. Note that this property is determined by the principa part of L ony. Remark 1..1 If the operator L is eiptic, then we may assume that a eigenvaues of the matrix A(x) in (1.4) are positive: 0 < λ 1 (x) λ (x)... λ n (x) for a x Ω. The operator L (and the corresponding boundary vaue probem) is caed uniformy eiptic if inf{λ 1 (x) x Ω} > 0 hods. Note that if the operator L is eiptic with coefficients a ij C( Ω) then the function x A(x) is continuous on the compact set Ω and hence L is uniformy eiptic. Using n a ij (x)ξ i ξ j = ξ T A(x)ξ λ 1 (x)ξ T ξ, i,j=1 we obtain that the operator L is uniformy eiptic if and ony if there exists a constant α 0 > 0 such that for a ξ R n, we have n a ij (x)ξ i ξ j α 0 ξ T ξ for a x Ω. i,j=1 We obtain a boundary vaue probem when we combine the partia differentia equation in (1.3) with certain boundary conditions for the unknown function u. For ease we restrict ourseves to probems with Dirichet boundary conditions, i.e., we impose : u = g on Ω, with g a given function on Ω. Other types of boundary conditions are the so caed Neumann boundary conditon, i.e., a condition on the norma derivative u n on Ω, and the mixed boundary condition which is a inear combination of a Dirichet and a Neumann boundary condition. Summarizing, we consider a inear second order Dirichet boundary vaue probem in Ω R n : Lu = n i,j=1 a ij u + x i x j n i=1 b i u x i + cu = f in Ω, (1.5a) u = g on Ω, (1.5b) where (a ij (x)) 1 i,j n is such that the probem is eiptic. A soution u of (1.5) is caed a cassica soution if u C (Ω) C( Ω). The functions (a ij (x)) 1 i,j n, (b i (x)) 1 i n and c(x) are caed the coefficients of the operator L. 1

13 1.. Exampes We assume n =, i.e. a probem with two independent variabes, say x 1 = x and x = y. Then the differentia operator is given by u Lu = a 11 x + a u 1 x y + a u y + b u 1 x + b u y + cu. In this case we have λ 1 (x)λ (x) = det(a(x)) and the eipticity condition can be formuated as a 11 (x,y)a (x,y) a 1 (x,y) > 0, (x,y) Ω. Exampes of eiptic equations are the Lapace equation the Poisson equation (cf. Poisson [7]) the reaction-diffusion equation and the convection-diffusion equation u := u x + u = 0 in Ω, y u = f in Ω, (1.6) u + cu = f in Ω, (1.7) u ε u + b 1 x + b u = f in Ω, ε > 0. (1.8) y If we add Dirichet boundary conditions to the Poisson equation in (1.6), we obtain the cassica Dirichet probem for Poisson s equation: { u = f in Ω, (1.9) u = g on Ω. Remark 1.. We briefy comment on the convection-diffusion equation in (1.8). If ε/b 1 1 or ε/b 1 (in a part of the domain) then the diffusion term ε u can be seen as a u perturbation of the convection term b 1 x + b u y (in a part of the domain). The convectiondiffusion equation is of eiptic type. However, for ε = 0 we obtain the so-caed reduced equation which is of hyperboic type. In view of this the convection-diffusion equation with ε/b 1 1 or ε/b 1 is caed a singuary perturbed equation. The fact that the eiptic equation (1.8) is then in some sense cose to a hyperboic equation resuts in some specia phenomena, that do not occur in diffusion dominated probems (as (1.9)). For exampe, in a convection dominated probem (e.g., an equation as (1.8) with ε 1 and b i = 1, i = 1, ) the soution u shows a behaviour in which most of the information is transported in certain directions ( streamines ). So we observe a behaviour as in the hyperboic probem (ε = 0), in which the soution satisfies an ordinary differentia equation aong each characteristic. Another phenomenon is the occurrence of boundary ayers. If we combine the equation in (1.8) with Dirichet boundary conditions on Ω then in genera these boundary conditions are not appropriate for the hyperboic probem (ε = 0). As a resut, if ε/b 1 1 or ε/b 1 we often observe that on a part of the boundary (corresponding to the outfow boundary in the hyperboic probem) there is a sma neighbourhood in which the soution u varies very rapidy. Such a neighbourhood is caed a boundary ayer. 13

14 For a detaied anaysis of singuary perturbed convection-diffusion equations we refer to Roos et a [76]. An iustration of the two phenomena described above is given in Section??. Finay we note that for the numerica soution of a probem with a singuary perturbed equation specia toos are needed, both with respect to the discretization of the probem and the iterative sover for the discrete probem Existence, uniqueness, reguarity For the eiptic boundary vaue probems introduced above, a first important topic that shoud be addressed concerns the existence and uniqueness of a soution. If a unique soution exists then another issue is the smoothness of the soution and how this smoothness depends on the data (source term, boundary condition, coefficients). Such smoothness resuts are caed reguarity properties of the probem. The topic of existence, uniqueness and reguarity has been, and sti is, the subject of many mathematica studies. We wi not treat these topics here. We ony give a few references to standard books in this fied: Gibarg and Trudinger [39], Miranda [64], Lions and Magenes [60], Hackbusch [45], [47]. We note that for the cassica formuation of an eiptic boundary vaue probem it is often rather hard to estabish satisfactory resuts on existence, uniqueness or reguarity. In Section.5 we wi discuss the variationa (or weak) formuation of eiptic boundary vaue probems. In that setting, additiona toos for the anaysis of existence, uniqueness and reguarity are avaiabe and (much) more resuts are known. Exampe 1..3 The reaction-diffusion equation can be used to show that a soution of an eiptic Dirichet boundary vaue probem as in (1.5) need not be unique. Consider the probem in (1.7) on Ω = (0,1), with f = 0 and c(x,y) = (µπ) (νπ),µ,ν N, combined with zero Dirichet boundary conditions. Then both u(x, y) 0 and u(x, y) = sin(µπx) sin(νπy) are soutions of this boundary vaue probem. Exampe 1..4 Even for very simpe eiptic probems a cassica soution may not exist. Consider a u x = 1 in (0,1), u(0) = u(1) = 0, with a(x) = 1 for 0 x 0.5 and a(x) = for 0.5 < x 1. Ceary the second derivative of a soution u of this probem cannot be continuous at x = 0.5. We present a typica resut from the iterature on existence and uniqueness of a cassica soution. For this we need a certain condition on Ω. The domain Ω is said to satisfy the exterior sphere condition if for every x 0 Ω there exists a ba B such that B Ω = x 0. Note that this condition is fufied, for exampe, if Ω is convex or if Ω C,0. 14

15 Theorem 1..5 ([39], Theorem 6.13) Consider the boundary vaue probem (1.5) and assume that (i) L is uniformy eiptic, (ii) Ω satisfies the exterior sphere condition, (iii) the coefficients of L and the function f beong to C 0,λ (Ω), λ (0,1), (iv) c 0 hods, (v) the boundary data are continuous : g C( Ω). Then the probem (1.5) has a unique cassica soution u C,λ (Ω) C( Ω). With respect to reguarity of the soution it is important to distinguish between interior smoothness (i.e., in Ω) and goba smoothness (i.e., in Ω). A typica resut on interior smoothness is given in the next theorem: Theorem 1..6 ([39], Theorem 6.17) Let u C (Ω) be a soution of (1.5). Suppose that L is eiptic and that there are k N, λ (0,1) such that the coefficients of L and the function f are in C k,λ (Ω). Then u C k+,λ (Ω) hods. If the coefficients and f are in C (Ω), then u C (Ω). This resut shows that the interior reguarity depends on the smoothness of the coefficients and of the right handside f, but does not depend on the smoothness of the boundary (data). A resut on goba reguarity is given in: Theorem 1..7 ([39], Theorem 6.19) Let u C (Ω) C( Ω) be a cassica soution of (1.5). Suppose that L is uniformy eiptic and that there are k N, λ (0,1) such that the coefficients of L and the function f are in C k,λ ( Ω), Ω C k+,λ. Assume that g can be extended on Ω such that g C k+,λ ( Ω). Then u C k+,λ ( Ω) hods. For a goba reguarity resut as in the previous theorem to hod, the smoothness of the boundary (data) is important. In practice one often has a domain with a boundary consisting of the union of straight ines (in D) or panes (3D). Then the previous theorem does not appy and the goba reguarity of the soution can be rather ow as is shown in the next exampe. Exampe 1..8 [from [47], p.13] We consider (1.9) with Ω = (0,1) (0,1), f 0, g(x,y) = x (so g C( Ω), g C ( Ω)). Then Theorem 1..5 guarantees the existence of a unique cassica soution u C (Ω) C( Ω). However, u is not an eement of C ( Ω). Proof. Assume that u C ( Ω) hods. From this and u = 0 in Ω it foows that u = 0 in Ω hods. From u = g = x on Ω we get u xx (x,0) = for x [0,1] and u yy (0,y) = 0 for y [0,1]. It foows that u(0,0) = must hod, which yieds a contradiction. 1.3 The Stokes equations The n-dimensiona Navier-Stokes equations mode the motion of an incompressibe viscous medium. It can be derived using basic principes from continuum mechanics (cf. [43]). The unknowns are the veocity fied u(x) = (u 1 (x),...,u n (x)) and the pressure p(x), x Ω. If one 15

16 considers a steady-state situation then these Navier-Stokes equations, in dimensioness quantities, are as foows: ν u i + n j=1 u j u i x j + p x i = f i in Ω, 1 i n, (1.10a) n j=1 u j x j = 0 in Ω, (1.10b) with ν > 0 a parameter that is reated to the viscosity of the medium. Using the notation u := ( u 1,..., u n ) T, divu := n u j j=1 x j, f = (f 1,...,f n ) T, we obtain the more compact formuation ν u + (u )u + p = f in Ω, (1.11a) div u = 0 in Ω. (1.11b) Note that the pressure p is determined ony up to a constant by these Navier-Stokes equations. The probem has to be competed with suitabe boundary conditions. One simpe possibiity is to take homogeneous Dirichet boundary conditions for u, i.e., u = 0 on Ω. If in the Navier- Stokes equations the noninear convection term (u )u is negected, which can be justified in situations where the viscosity parameter ν is arge, one obtains the Stokes equations. From a simpe rescaing argument (repace u by 1 νu) it foows that without oss of generaity in the Stokes equations we can assume ν = 1. Summarizing, we obtain the foowing Stokes probem: u + p = f in Ω, (1.1a) div u = 0 in Ω, (1.1b) u = 0 on Ω. (1.1c) This is a stationary boundary vaue probem, consisting of n + 1 couped partia differentia equations for the unknowns (u 1,...,u n ) and p. In [] the notion of eipticity is generaized to systems of partia differentia equations. It can be shown (cf. [, 47]) that the Stokes equations indeed form an eiptic system. We do not discuss existence and uniqueness of a cassica soution of the Stokes probem. In chapter we discuss the varitiona formuation of the Stokes probem. For this formuation the issue of existence, uniqueness and reguarity of a soution is treated in section.6. 16

17 Chapter Weak formuation.1 Introduction For soving a boundary vaue probem it can be (very) advantageous to consider a generaization of the cassica probem formuation, in which arger function spaces are used and a weaker soution (expained beow) is aowed. This resuts in the variationa formuation (aso caed weak formuation) of a boundary vaue probem. In this section we consider an introductory exampe which iustrates that even for a very simpe boundary vaue probem the choice of an appropriate soution space is an important issue. This exampe aso serves as a motivation for the introduction of the Soboev spaces in section.. Consider the foowing eiptic two-point boundary vaue probem: (au ) = 1 in (0,1), u(0) = u(1) = 0. (.1a) (.1b) We assume that the coefficient a is an eement of C 1 ([0,1]) and that a(x) > 0 hods for a x [0,1]. This probem then has a unique soution in the space This soution is given by u(x) = x 0 V 1 = {v C ([0,1]) v(0) = v(1) = 0 }. t + c a(t) ( 1 dt, c := 0 t )( 1 a(t) dt 0 1 a(t) dt ) 1, which may be checked by substitution in (.1). If one mutipies the equation (.1a) by an arbitrary function v V 1, integrates both the eft and right handside and then appies partia integration one can show that u V 1 is the soution of (.1) if and ony if 1 0 a(x)u (x)v (x)dx = 1 0 v(x)dx for a v V 1. (.) This variationa probem can be reformuated as a minimization probem. For this we introduce the notion of a biinear form. 17

18 Definition.1.1 Let X be a vector space. A mapping k : X X R is caed a biinear form if for arbitrary α, β R and u, v, w X the foowing hods: k(αu + βv,w) = αk(u,w) + β k(v,w), k(u,αv + βw) = αk(u,v) + β k(u,w). The biinear form is symmetric if k(u,v) = k(v,u) hods for a u,v X. Lemma.1. Let X be a vector space and k : X X R a symmetric biinear form which is positive, i.e., k(v,v) > 0 for a v X, v 0. Let f : X R be a inear functiona. Define J : X R by J(v) = 1 k(v,v) f(v). Then J(u) < J(v) for a v X, v u, hods if and ony if Moreover, there exists at most one minimizer u of J( ). Proof. Take u,w X, t R. Note that k(u,v) = f(v) for a v X. (.3) J(u + tw) J(u) = t ( k(u,w) f(w) ) + 1 t k(w,w) =: g(t;u,w).. If J(u) < J(v) for a v X, v u, then the function t g(t;u,w) must be stricty positive for a w X \ {0} and t R \ {0}. It foows that k(u,w) f(w) = 0 for a w X.. From (.3) it foows that J(u + tw) J(u) = 1 t k(w,w) for a w X, t R. For v u, w = v u, t = 1 this yieds J(v) J(u) = 1 k(v u,v u) > 0. We finay prove the uniqueness of the minimizer. Assume that k(u i,v) = f(v) for a v X and for i = 1,. It foows that k(u 1 u,v) = 0 for a v X. For the choice v = u 1 u we get k(u 1 u,u 1 u ) = 0. From the property that the biinear form k(, ) is positive we concude u 1 = u. Note that in this emma we do not caim existence of a soution. For the minimizer u (if it exists) the reation J(v) J(u) = 1 k(v u,v u) for a v X (.4) hods. We now return to the exampe and take X = V 1, k(u,v) = 1 0 a(x)u (x)v (x)dx, f(v) = 1 0 v(x) dx. Note that a assumptions of emma.1. are fufied. It then foows that the unique soution of (.1) (or, equivaenty, of (.)) is aso the unique minimizer of the functiona J(v) = 1 0 [ 1 a(x)v (x) v(x)]dx. (.5) We consider a case in which the coefficient a is ony piecewise continuous (and not differentiabe at a x (0,1)). Then the probem in (.1) is not we-defined. However, the definitions of the 18

19 biinear form k(, ) and of the functiona J( ) sti make sense. We now anayze a minimization probem with a functiona as in (.5) in which the coefficient a is piecewise constant: { 1 if x [0, 1 a(x) = ] if x ( 1,1]. We show that for this probem the choice of an appropriate soution space is a deicate issue. Note that due to emma.1. the minimization probem in X = V 1 has a corresponding equivaent variationa formuation as in (.3). With our choice of the coefficient a the functiona J( ) takes the form J(v) := 1 0 [ 1 v (x) v(x)]dx [v (x) v(x)]dx. (.6) This functiona is we-defined on the space V 1. The functiona J, however, is aso we-defined if v is ony differentiabe and even if we aow v to be nondifferentiabe at x = 1. We introduce the spaces V = {v C 1 ([0,1]) v(0) = v(1) = 0 }, V 3 = {v C 1 ([0, 1 ]) C1 ([ 1,1]) C([0,1]) v(0) = v(1) = 0 }, V 4 = {v C 1 ([0, 1 ]) C1 ([ 1,1]) v(0) = v(1) = 0 }. Note that V 1 V V 3 V 4 and that on a these spaces the functiona J( ) is we-defined. Moreover, with X = V i, i = 1,...,4, and k(u,v) = 1 0 u (x)v (x)dx u (x)v (x)dx, f(v) = 1 0 v(x) dx (.7) a assumptions of emma.1. are fufied. We define a (natura) norm on these spaces: w := 1 0 w (x) dx w (x) dx. (.8) One easiy checks that this indeed defines a norm on the space V 4 and thus aso on the subspaces V i, i = 1,,3. Furthermore, this norm is induced by the scaar product on V 4, and (w,v) 1 := 1 0 w (x)v (x)dx w (x)v (x)dx (.9) 1 w k(w,w) w for a w V 4 (.10) hods. We show that in the space V 3 the minimization probem has a unique soution. Lemma.1.3 The probem min v V3 J(v) has a unique soution u given by u(x) = { 1 x x if 0 x 1, 1 4 x x if x 1. (.11) 19

20 Proof. Note that u V 3 and even u C ([0, 1 ]) C ([ 1,1]). We use the notation u L (1 ) = im x 1 u (x) and simiary for u R (1 ). We appy emma.1. with X = V 3. For arbitrary v V 3 we have k(u,v) f(v) = 1 0 u (x)v (x) v(x)dx + 1 = u L( 1 1 )v(1 ) (u (x) + 1)v(x)dx 0 u R( 1 1 )v(1 ) (u (x) + 1)v(x)dx. 1 1 u (x)v (x) v(x)dx (.1) Due to u (x) = 1 on [0, 1 ], u (x) = 1 on [1,1] and u L (1 ) u R (1 ) = 0 we obtain k(u,v) = f(v) for a v V 3. From emma.1. we concude that u is the unique minimizer in V 3. Thus with X = V 3 a minimizer u exists and the reation (.4) takes the form J(v) J(u) = 1 k(v u,v u), with k(, ) as in (.7). Due to (.10) the norm can be used as a measure for the distance from the minimum (i.e. J(v) J(u)): 1 4 v u J(v) J(u) 1 v u. (.13) Before we turn to the the minimization probems in the spaces V 1 and V we first present a usefu emma. Lemma.1.4 Define W := {v C ([0,1]) v(0) = v(1) = 0 }. For every u V 3 there is a sequence (u n ) n 1 in W such that im u n u = 0. (.14) n Proof. Take u V 3 and define û(x) := u (x) for a x [0,1], x 1, û(1 ) a fixed arbitrary vaue and û( x) = û(x) for a x [0,1]. Then û is even and û L ( ( 1,1) ). From Fourier anaysis it foows that there is a sequence such that û n (x) = n a k cos(kπx), n N, k=0 1 im û û n (û(x) n L = im ûn (x) ) dx = 0. n û(x)dx = Note that due to the fact that u is continuous and u(0) = u(1) = 0 we get a 0 = u (x)dx u (x)dx = 0. Define u n (x) = n a k k=1 kπ sin(kπx) For n 1. Then u n W, u n = û n and u u n û û n L hods. Thus it foows that im n u n u = 0. 0

21 Lemma.1.5 Let u V 3 be given by (.11). For i = 1, the foowing hods: inf J(v) = min J(v) = J(u). (.15) v V i v V 3 Proof. Take i = 1 or i =. I := inf v Vi J(v) is defined as the greatest ower bound of J(v) for v V i. From V 3 V i it foows that J(u) I hods. Suppose that J(u) < I hods, i.e. we have δ := I J(u) > 0. Due to W V i and emma.1.4 there is a sequence (u n ) n 1 in V i such that im n u u n = 0 hods. Using (.13) we obtain J(u n ) = J(u) + (J(u n ) J(u)) I δ + 1 u n u. So for n sufficienty arge we have J(u n ) < I, which on the other hand is not possibe because I is a ower bound of J(v) for v V i. We concude that J(u) = I hods. The resut in this emma shows that the infimum of J(v) for v V is equa to J(u) and thus, using (.13), it foows that the minimizer u V 3 can be approximated to any accuracy, measured in the norm, by eements from the smaer space V. The question arises why in the minimization probem the space V 3 is used and not the seemingy more natura space V. The answer to this question is formuated in the foowing emma. Lemma.1.6 There does not exist u V such that J(u) J(v) for a v V. Proof. Suppose that such a minimizer, say ū V, exists. From emma.1.5 we then obtain J(ū) = min v V J(v) = inf v V J(v) = J(u) with u as in (.11) the minimizer in V 3. Note that ū V 3. From emma.1. it foows that the minimizer in V 3 is unique and thus ū = u must hod. But then ū = u / V, which yieds a contradiction. The same arguments as in the proof of this emma can be used to show that in the smaer space V 1 there aso does not exist a minimizer. Based on these resuts the function u V 3 is caed the weak soution of the minimization probem in V. From (.15) we see that for soving the minimization probem in the space V, in the sense that one wants to compute inf v V J(v), it is natura to consider the minimization probem in the arger space V 3. We now consider the even arger space V 4 and show that the minimization probem sti makes sense (i.e. has a unique soution). However, the minimum vaue does not equa inf v V J(v). Lemma.1.7 The probem min v V4 J(v) has a unique soution ū given by { 1 ū(x) = x(x 1) if 0 x 1, 1 4 x(x 1) if 1 < x 1. Proof. Note that ū V 4 hods. We appy emma.1.. Reca the reation (.1): k(ū,v) f(v) = ū L( 1 1 )v(1 ) (ū (x) + 1)v(x)dx 0 ū R( 1 1 )v(1 ) (ū (x) + 1)v(x)dx. 1 1

22 From ū (x) = 1 on [0, 1 ], ū (x) = 1 on [1,1] and ū L (1 ) = ū R (1 ) = 0 it foows that k(ū,v) = f(v) for a v V 4. We concude that ū is the unique minimizer in V 4. A straightforward cacuation yieds the foowing vaues for the minima of the functiona J( ) in V 3 and in V 4, respectivey: J(u) = , J(ū) = 1 3. From this we see that, opposite to u, we shoud not ca ū a weak soution of the minimization probem in V, because for ū we have J(ū) < inf v V J(v). We concude the discussion of this exampe with a few remarks on important issues that pay an important roe in the remainder of this book 1) Both the theoretica anaysis and the numerica soution methods treated in this book heaviy rey on the varationa formuation of the eiptic boundary vaue probem (as, for exampe, in (.)). In section.3 genera resuts on existence, uniquess and stabiity of variationa probems are presented. In the sections.5 and.6 these are appied to variationa formuations of eiptic boundary vaue probems. The finite eement discretization method treated in chapter 3 is based on the variationa formuation of the boundary vaue probem. The derivation of the conjugate gradient (CG) iterative method, discussed in chapter 7, is based on the assumption that the given (discrete) probem can be formuated as a minimization probem with a functiona J very simiar to the one in emma.1.. ) The biinear form used in the weak formuation often has properties simiar to those of an inner product, cf. (.7), (.9), (.10). To take advantage of this one shoud formuate the probem in an inner product space. Then the structure of the space (inner product) fits nicey to the variationa probem. 3) To guarantee that a weak soution actuay ies in the space one shoud use a space that is arge enough but not too arge. This can be reaized by competion of the space in which the probem is formuated. The concept of competion is expained in section... The conditions discussed in the remarks ) and 3) ead to Hibert spaces, i.e. inner product spaces that are compete. The Hibert spaces that are appropriate for eiptic boundary vaue probems are the Soboev spaces. These are treated in section... Soboev spaces The Höder spaces C k,λ ( Ω) that are used in the cassica formuation of eiptic boundary vaue probems in chapter 1 are Banach spaces but not Hibert spaces. In this section we introduce Soboev spaces. A Soboev spaces are Banach spaces. Some of these are Hibert spaces. In our treatment of eiptic boundary vaue probems we ony need these Hibert spaces and thus we restrict ourseves to the presentation of this subset of Soboev Hibert spaces. A very genera treatment on Soboev spaces is given in [1].

23 ..1 The spaces W m (Ω) based on weak derivatives We take u C 1 (Ω) and φ C0 (Ω). Since φ vanishes identicay outside some compact subset of Ω, one obtains by partia integration in the variabe x j : u(x) φ(x)dx = u(x) φ(x) dx x j x j Ω and thus D α u(x)φ(x)dx = u(x)d α φ(x)dx, α = 1, Ω Ω hods. Repeated appication of this resut yieds the fundamenta Green s formua D α u(x)φ(x)dx = ( 1) α u(x)d α φ(x)dx, Ω for a φ C 0 (Ω), u Ck (Ω), k = 1,,... and α k. Based on this formua we introduce the notion of a weak derivative: Ω Ω (.16) Definition..1 Consider u L (Ω) and α > 0. If there exists v L (Ω) such that v(x)φ(x)dx = ( 1) α u(x)d α φ(x)dx for a φ C0 (Ω) (.17) Ω then v is caed the αth weak derivative of u and is denoted by D α u := v. Two eementary resuts are given in the next emma. Ω Lemma.. If for u L (Ω) the αth weak derivative exists then it is unique (in the usua Lebesgue sense). If u C k (Ω) then for 0 < α k the αth weak derivative and the cassica αth derivative coincide. Proof. The second statement foows from the first one and Green s formua (.16). We now prove the uniqueness. Assume that v i L (Ω), i = 1,, both satisfy (.17). Then it foows that ( v1 (x) v (x) ) φ(x)dx = v 1 v,φ L = 0 for a φ C0 (Ω) Ω Since C 0 (Ω) is dense in L (Ω) this impies that v 1 v,φ L = 0 for a φ L (Ω) and thus v 1 v = 0 (a.e.). Remark..3 As a warning we note that if the cassica derivative of u, say D α u, exists amost everywhere in Ω and D α u L (Ω), this does not guarantee the existence of the αth weak derivative. This is shown by the foowing exampe: Ω = ( 1,1), u(x) = 0 if x < 0, u(x) = 1 if x 0. The cassica first derivative of u on Ω \ {0} is u (x) = 0. However, the weak first derivative of u as defined in..1 does not exist. A further noticabe observation is the foowing: If u C (Ω) C(Ω) then u does not aways have a first weak derivative. This is shown by the exampe Ω = (0,1), u(x) = x. The ony candidate for the first weak derivative of u is v(x) = 1 x. However, v / L (Ω). 3

24 The Soboev space W m (Ω), m = 1,,..., consists of a functions in L (Ω) for which a αth weak derivatives with α m exist: W m (Ω) := {u L (Ω) D α u exists for a 0 < α m } (.18) Remark..4 By definition, for u W m (Ω), Green s formua D α u(x)φ(x)dx = ( 1) α u(x)d α φ(x)dx, for a φ C0 hods. Ω Ω (Ω), α m, For m = 0 we define W 0 (Ω) := L (Ω). In W m (Ω) a natura inner product and corresponding norm are defined by u,v m := D α u,d α v L, u m := u,u 1 m, u,v W m (Ω) (.19) α m It is easy to verify, that, m defines an inner product on W m (Ω). We now formuate a main resut: Theorem..5 The space (W m (Ω),, m ) is a Hibert space. Proof. We must show that the space W m (Ω) with the norm m is compete. For m = 0 this is trivia. We consider m 1. First note that for v W m (Ω): v m = D α v L (.0) α m Let (u k ) k 1 be a Cauchy sequence in W m (Ω). From (.0) it foows that if u k u m ε then D α u k D α u L ε for a 0 α m. Hence, (D α u k ) k 1 is a Cauchy sequence in L (Ω) for a α m. Since L (Ω) is compete it foows that there exists a unique u (α) L (Ω) with im k D α u k = u (α) in L (Ω). For α = 0 this yieds im k u k = u (0) in L (Ω). For 0 < α m and arbitrary φ C0 (Ω) we obtain u (α),φ L = im k Dα u k,φ L = im k ( 1) α u k,d α φ L = ( 1) α u (0),D α φ L (.1) From this it foows that u (α) L (Ω) is the αth weak derivative of u (0). We concude that u (0) W m (Ω) and im u k u (0) m = im k k = α m α m D α u k D α u (0) L im k Dα u k u (α) L = 0 This shows that the Cauchy sequence (u k ) k 1 in W m (Ω) has its imit point in W m (Ω) and thus this space is compete. 4

25 Simiar constructions can be appied if we repace the Hibert space L (Ω) by the Banach space L p (Ω), 1 p < of measurabe functions for which u p := ( Ω u(x) p dx) 1/p is bounded. This resuts in Soboev spaces which are usuay denoted by Wp m (Ω). For notationa simpicity we deeted the index p = in our presentation. For p the Soboev space Wp m (Ω) is a Banach space but not a Hibert space. In this book we ony need the Soboev space with p = as defined in (.18). For p we refer to the iterature, e.g. [1]... The spaces H m (Ω) based on competion In this section we introduce the Soboev spaces using a different technique, namey based on the concept of competion. We reca a basic resut (cf. Appendix A.1). Lemma..6 Let (Z, ) be a normed space. Then there exists a Banach space (X, ) such that Z X, x = x for a x Z, and Z is dense in X. The space X is caed the competion of Z. This space is unique, except for isometric (i.e., norm preserving) isomorphisms. Here we consider the function space Z m := {u C (Ω) u m < }, (.) endowed with the norm m, as defined in (.19), i.e, we want to construct the competion of (Z m, m ). For m = 0 this resuts in L (Ω), since C (Ω) is dense in L (Ω). Hence, we ony consider m 1. Note that Z m W m (Ω) and that this embedding is continuous. One can appy the genera resut of emma..6 which then defines the competion of the space Z m. However, here we want to present a more constructive approach which reveas some interesting reations between this competion and the space W m (Ω). First, note that due to (.0) a Cauchy sequence (u k ) k 1 in Z m is aso a Cauchy sequence in L (Ω) and thus to every such sequence there corresponds a unique u L (Ω) with im k u k u L = 0. The space V m Z m is defined as foows: V m := {u L (Ω) im k u k u L = 0 for a Cauchy sequence (u k ) k 1 in Z m } One easiy verifies that V m is a vector space. Lemma..7 V m is the cosure of Z m in the space W m (Ω). Proof. Take u V m and et (u k ) k 1 be a Cauchy sequence in Z m with im k u k u L = 0. From (.0) it foows that (D α u k ) k 1, 0 < α m are Cauchy sequences in L (Ω). Let u (α) := im k D α u k in L (Ω). As in (.1) one shows that u (α) is the αth weak derivative D α u of u. Using D α u = im k D α u k in L (Ω), for 0 < α m, we get im u k u m = im k k α m D α u k D α u L = 0 Since (u k ) k 1 is a sequence in Z m we have shown that V m is the cosure of Z m in W m (Ω). On the space V m we can take the same inner product (and corresponding norm) as used in the space W m (Ω) (cf. (.19)). From emma..7 and the fact that in the space Z m the norm is the same as the norm of W m (Ω) it foows that (V m, m ) is the competion of (Z m, m ). 5

26 Since the norm m is induced by an inner product we have that (V m,, m ) is a Hibert space. This defines the Soboev space It is cear from emma..7 that H m (Ω) := (V m,, m ) = competion of (Z m,, m ) hods. A fundamenta resut is the foowing: H m (Ω) W m (Ω) Theorem..8 The equaity H m (Ω) = W m (Ω) hods. Proof. The first proof of this resut was presented in [63]. A proof can aso be found in [1, 65]. We see that the construction using weak derivatives (space W m (Ω)) and the one based on competion (space H m (Ω)) resut in the same Soboev space. In the remainder we wi ony use the notation H m (Ω). The resut in theorem..8 hods for arbitrary open sets Ω in R n. If the domain satisfies certain very mid smoothnes conditions (it suffices to have assumption 1.1.3) one can prove a somewhat stronger resut, that we wi need further on: Theorem..9 Let H m (Ω) be the competion of the space (C ( Ω),, m ). Then hods. Proof. We refer to [1]. H m (Ω) = H m (Ω) = W m (Ω) Note that C ( Ω) Z m and thus H m (Ω) resuts from the competion of a smaer space than H m (Ω). Remark..10 If assumption is not satisfied then it may happen that H m (Ω) W m (Ω) hods. Consider the exampe Ω = {(x,y) R x ( 1,0) (0,1), y (0,1) } and take u(x,y) = 1 if x > 0, u(x,y) = 0 if x < 0. Then D (1,0) u = D (0,1) u = 0 on Ω and thus u W 1 (Ω). However, one can verify that there does not exist a sequence (φ k ) k 1 in C 1 ( Ω) such that u φ k 1 = u φ k L + D α φ k L 0 for k α =1 Hence, C 1 ( Ω) is not dense in W 1 (Ω), i.e., H1 (Ω) W 1 (Ω). The equaity H 1 (Ω) = W 1 (Ω), however, does hod. The competion can aso be defined if in (.) the space C (Ω) is repaced by the smaer space C0 (Ω). This yieds another cass of important Soboev spaces: H m 0 (Ω) competion of the space (C 0 (Ω),, m) (.3) The space H m 0 (Ω) is a Hibert space that is in genera stricty smaer than Hm (Ω). 6

27 Remark..11 In genera we have H0 1(Ω) H1 (Ω). Consider, as a simpe exampe, Ω = (0,1), u(x) = x. Then u H 1 (Ω) but for arbitrary φ C0 (Ω) we have u φ 1 = u φ L + u φ L ( u (x) φ (x) ) 1 dx = 1 φ (x) + φ (x) dx 1 0 φ (x)dx = 1 ( φ(1) φ(0) ) = 1 Hence u / H 1 0 (Ω) = C 0 (Ω) 1. The technique of competion can aso be appied if instead of m one uses the norm u m,p = ( α m Dα u p ) 1/p, 1 p <. This resuts in Soboev spaces denoted by H m p (Ω). For p = we have H m (Ω) = Hm (Ω). For p these spaces are Banach spaces but not Hibert spaces. A resut as in theorem..8 aso hods for p : H m p (Ω) = W m p (Ω). We now formuate a resut on a certain cass of piecewise smooth functions which form a subset of the Soboev space H m (Ω). This subset pays an important roe in the finite eement method that wi be presented in chapter 3. Theorem..1 Assume that Ω can be partitioned as Ω = N i=1 Ω i, with Ω i Ω j = for a i j and for a Ω i the assumption is fufied. For m N, m 1, define For u V m the foowing hods: V m = {u L (Ω) u Ωi C m ( Ω i ) for a i = 1,...,N } u H m (Ω) u C m 1 ( Ω) Proof. First we need some notation. Let Γ i := Ω i. The outward unit norma on Γ i is denoted by n (i). Let Γ i := Γ i Γ (= Γ i ) and γ int denotes the set of a those intersections Γ i with meas n 1 (Γ i ) > 0 (in D with trianges: intersections by sides are taken into account but intersections by vertices not). Simiary, Γ i0 := Γ i Ω and γ b the set of a Γ i0 with meas n 1 (Γ i0 ) > 0. For Γ i γ int et n (i) be the unit norma pointing outward from Ω i (thus n (i) = n (i). Finay, for u V 1 et 0 [u] i = im t 0 ( u(x + tn (i) ) u(x + tn (i) ) ), x Γ i γ int be the jump of u across Γ i. We now consider the case m = 1, i.e., u V 1. Let v L (Ω) be given by v(x) = u(x) x k for x Ω i, i = 1,...,N. For arbitrary φ C0 (Ω) we have Ω u(x) φ(x) x k dx = = N u(x) φ(x) i=1 Ω i x k N i=1 = Ω dx Ω i v(x)φ(x)dx + v(x)φ(x) dx + 7 N i=1 N i=1 Γ i u(x)φ(x)n (i) k dx u(x)φ(x)n (i) k Γ i dx (.4)

28 For the ast term in this expression we have N i=1 Γ i u(x)φ(x)n (i) k dx = Γ i γ int + Γ i0 γ b [u] i φ(x)n (i) k Γ i dx Γ i0 u(x)φ(x)n (i) k dx =: R int + R b We have R b = 0 because φ(x) = 0 on Ω. If u H 1 (Ω) hods then the weak derivative u x k must be equa to v (for a k = 1,...,n). From Ω u(x) φ(x) x k u(x) dx = φ(x)dx = Ω x k Ω v(x)φ(x)dx, φ C 0 (Ω), it foows that R int = 0 must hod for a φ C0 (Ω). This impies that the jump of u across Γ i is zero and thus u C( Ω) hods. Conversey, if u C( Ω) then R int = 0 and from the reation (.4) it foows that the weak derivative u x k exists. Since k is arbitrary we concude u H 1 (Ω). This competes the proof for the case m = 1. For m > 1 we use an induction argument. Assume that the statement hods for m. We consider m + 1. Take u V m+1 and assume that u H m+1 (Ω) hods. Take an arbitrary muti-index α with α m 1. Cassica derivatives wi be denoted by ˆD β and weak ones by D β (with β a muti-index as α). From the induction hypothesis we obtain w := ˆD α u C( Ω). From u H m+1 (Ω) it foows that D β w H 1 (Ω) for β 1. Furthermore, for these β vaues we aso have, due to u V m+1 that D β w = ˆD β w C 1 ( Ω i ) for i = 1,...,N. From the resut for m = 1 it now foows that D β w C( Ω) for β 1. Hence, ˆDβ w is continuous across the interna interfaces Γ i and thus ˆD β w C( Ω) hods. We concude that ˆD α ˆDβ u C( Ω) for a α m 1, β 1, i.e., u C m ( Ω). Conversey, if u V m+1 and u C m ( Ω) then ˆD α u C( Ω) for α m. From the resut for m = 1 it foows that ˆD α u H 1 (Ω) for a α m and thus u H m+1 (Ω) hods...3 Properties of Soboev spaces There is an extensive iterature on the theory of Soboev spaces, see for exampe, Adams [1], Marti [61], Nečas [65], Woka [99], At [3], and the references therein. In this section we coect a few resuts that wi be needed further on. A first important question concerns the smoothness of functions from H m (Ω) in the cassica (i.e., C k ( Ω)) sense. For exampe, one can show that if Ω R then a functions from H 1 (Ω) must be continuous on Ω. This, however, is not true for the two dimensiona case, as the foowing exampe shows: Exampe..13 In this exampe we show that functions in H 1 (Ω), with Ω R, are not necessariy continuous on Ω. With r := (x 1 + x )1 et B(0,α) := {(x 1,x ) R r < α } for α > 0. We take Ω = B(0, 1 ). Beow we aso use Ω δ := Ω \ B(0,δ) with 0 < δ < 1. On Ω we define the function u by u(0,0) := 0, u(x 1,x ) := n(n(1/r)) otherwise. Using poar coordinates one obtains 1 u(x) dx = im u(x) dx = π im [n(n(1/r))] r dr < δ 0 Ω δ δ 0 δ Ω 8

29 so u L (Ω) hods. Note that u C (Ω \ {0}). For the first derivatives we have Ω i=1, ( u(x) ) 1 dx = π im x i δ 0 δ 1 π r r dr = (n r) n It foows that the cassica first derivatives u x i exist a.e. on Ω and are eements of L (Ω). Note, however, remark..3. For arbitrary φ C 0 (Ω) we have, using Green s formua on Ω δ: Ω δ u(x) φ(x) x 1 dx = B(0,δ) u(x) u(x)φ(x)n x1 ds φ(x)dx. Ω δ x 1 Note that So we have im u(x)φ(x)n x1 ds im πδ φ n(n(1/δ)) = 0. δ 0 B(0,δ) δ 0 Ω u(x) φ(x) x 1 u(x) dx = φ(x)dx. Ω x 1 We concude that u x 1 is the generaized partia derivative with respect to the variabe x 1. The same argument yieds an anaogous resut for the derivative w.r.t. x. We concude that u H 1 (Ω). It is cear that u is not continuous on Ω. We now formuate an important genera resut which reates smoothness in the Soboov sense (weak derivatives) to cassica smoothness properties. For normed spaces X and Y a inear operator I : X Y is caed a continuous embedding if I is continuous and injective. Such a continuous embedding is denoted by X Y. Furthermore, usuay for x X the corresponding eement Ix Y is denoted by x, too (X Y is formay repaced by X Y ). Theorem..14 If m n > k (reca: Ω Rn ) then there exist continuous embeddings H m (Ω) C k ( Ω) H m 0 (Ω) Ck ( Ω). (.5a) (.5b) Proof. Given in [1] It is trivia that for m 0 there are continuous embeddings H m+1 (Ω) H m (Ω) and H0 m+1 (Ω) H0 m (Ω). In the next theorem a resut on compactness of embeddings (cf. Appendix A.1) is formuated. We reca that if X, Y are Banach spaces then a continuous embedding X Y is compact if and ony if each bounded sequence in X has a subsequence which is convergent in Y. 9

30 Theorem..15 The continuous embeddings are compact. H m+1 (Ω) H m (Ω) for m 0 (.6a) H0 m+1 (Ω) H0 m (Ω) for m 0 (.6b) H m (Ω) C k ( Ω) for m n > k (.6c) H m 0 (Ω) Ck ( Ω) for m n > k (.6d) Proof. We sketch the idea of the proof. In [1] it is shown that the embeddings H 1 (Ω) L (Ω), H 1 0 (Ω) L (Ω) are compact. This proves the resuts in (.6a) and (.6b) for m = 0. The resuts in (.6a) and (.6b) for m 1 can easiy be derived from this as foows. Let (u k ) k 1 be a bounded sequence in H m+1 (Ω) (m 1). Then (D α u k ) k 1 is a bounded sequence in H 1 (Ω) for α m. Thus this sequence has a subsequence (D α u k ) k 1 that converges in L (Ω). Hence, the subsequence (u k ) k 1 converges in H m (Ω). This proves the compactness of the embedding H m+1 (Ω) H m (Ω). The resut in (.6b) for m 1 can be shown in the same way. With a simiar shift argument one can easiy show that it suffices to prove the resuts in (.6c) and (.6d) for the case k = 0. The anaysis for the case k = 0 is based on the foowing genera resut (which is easy to prove). If X, Y, Z are normed spaces with continuous embeddings I 1 : X Y, I : Y Z and if at east one of these embeddings is compact then the continuous embedding I I 1 : X Z is compact. For m n > 0 there exist µ,λ (0,1) with 0 < λ < µ < m n. The foowing continuous embeddings exist: H m (Ω) C 0,µ ( Ω) C 0,λ ( Ω) C( Ω). In this sequence ony the first embedding is nontrivia. This one is proved in [1], theorem 5.4. Furthermore, from [1] theorem 1.31 it foows that the second embedding is compact. We concude that for m n > 0 the embedding Hm (Ω) C( Ω) is compact. This then yieds the resut in (.6c) for k = 0. The same ine of reasoning can be used to show that (.6d) hods. The resut in the foowing theorem is a basic inequaity that wi be used frequenty. Theorem..16 (Poincare-Friedrichs inequaity) There exists a constant C that depends ony on diam(ω) such that u L C D α u for a u H L 0(Ω). 1 α =1 Proof. Because C0 (Ω) is dense in H1 0 (Ω) it is sufficient to prove the inequaity for u C0 (Ω). Without oss of generaity we can assume that (0,...,0) Ω. Let a > 0 be such that Ω [ a,a] n =: E. Take u C0 (Ω) and extend this function by zero outside Ω. Note that u(x 1,...,x n ) = u( a,x,...,x n ) + 30 x1 a u(t,x,...,x n ) t dt