(bilinearity), a(u, v) M u V v V (continuity), a(v, v) m v 2 V (coercivity).

Similar documents
M A T H F A L L CORRECTION. Algebra I 1 4 / 1 0 / U N I V E R S I T Y O F T O R O N T O

b i u x i U a i j u x i u x j

Real Numbers R ) - LUB(B) may or may not belong to B. (Ex; B= { y: y = 1 x, - Note that A B LUB( A) LUB( B)

Linear Elliptic PDE s Elliptic partial differential equations frequently arise out of conservation statements of the form

Zeros of Polynomials

Math 61CM - Solutions to homework 3

A) is empty. B) is a finite set. C) can be a countably infinite set. D) can be an uncountable set.

Math Solutions to homework 6

PAPER : IIT-JAM 2010

Sequences and Series of Functions

, then cv V. Differential Equations Elements of Lineaer Algebra Name: Consider the differential equation. and y2 cos( kx)

Chapter 3 Inner Product Spaces. Hilbert Spaces

NBHM QUESTION 2007 Section 1 : Algebra Q1. Let G be a group of order n. Which of the following conditions imply that G is abelian?

Lecture Notes for Analysis Class

Bernoulli, Ramanujan, Toeplitz e le matrici triangolari

5.1. The Rayleigh s quotient. Definition 49. Let A = A be a self-adjoint matrix. quotient is the function. R(x) = x,ax, for x = 0.

A brief introduction to linear algebra

Physics 324, Fall Dirac Notation. These notes were produced by David Kaplan for Phys. 324 in Autumn 2001.

Chapter 6 Infinite Series

(A sequence also can be thought of as the list of function values attained for a function f :ℵ X, where f (n) = x n for n 1.) x 1 x N +k x N +4 x 3

Apply change-of-basis formula to rewrite x as a linear combination of eigenvectors v j.

62. Power series Definition 16. (Power series) Given a sequence {c n }, the series. c n x n = c 0 + c 1 x + c 2 x 2 + c 3 x 3 +

LECTURE 8: ORTHOGONALITY (CHAPTER 5 IN THE BOOK)

6.3 Testing Series With Positive Terms

Math 155 (Lecture 3)

It is often useful to approximate complicated functions using simpler ones. We consider the task of approximating a function by a polynomial.

6 Integers Modulo n. integer k can be written as k = qn + r, with q,r, 0 r b. So any integer.

Chapter 3. Strong convergence. 3.1 Definition of almost sure convergence

University of Colorado Denver Dept. Math. & Stat. Sciences Applied Analysis Preliminary Exam 13 January 2012, 10:00 am 2:00 pm. Good luck!

Abstract Vector Spaces. Abstract Vector Spaces

lim za n n = z lim a n n.

Definition 4.2. (a) A sequence {x n } in a Banach space X is a basis for X if. unique scalars a n (x) such that x = n. a n (x) x n. (4.

Infinite Sequences and Series

Machine Learning Theory Tübingen University, WS 2016/2017 Lecture 11

CHAPTER 5. Theory and Solution Using Matrix Techniques

Introduction to Optimization Techniques

Lecture 10: Bounded Linear Operators and Orthogonality in Hilbert Spaces

The Method of Least Squares. To understand least squares fitting of data.

1 Last time: similar and diagonalizable matrices

Iterative Techniques for Solving Ax b -(3.8). Assume that the system has a unique solution. Let x be the solution. Then x A 1 b.

Inverse Matrix. A meaning that matrix B is an inverse of matrix A.

REAL ANALYSIS II: PROBLEM SET 1 - SOLUTIONS

Theorem 3. A subset S of a topological space X is compact if and only if every open cover of S by open sets in X has a finite subcover.

3.2 Properties of Division 3.3 Zeros of Polynomials 3.4 Complex and Rational Zeros of Polynomials

Analytic Continuation

The picture in figure 1.1 helps us to see that the area represents the distance traveled. Figure 1: Area represents distance travelled

Measure and Measurable Functions

CALCULATION OF FIBONACCI VECTORS

Riesz-Fischer Sequences and Lower Frame Bounds

Review Problems 1. ICME and MS&E Refresher Course September 19, 2011 B = C = AB = A = A 2 = A 3... C 2 = C 3 = =

Chimica Inorganica 3

Lesson 10: Limits and Continuity

3. Z Transform. Recall that the Fourier transform (FT) of a DT signal xn [ ] is ( ) [ ] = In order for the FT to exist in the finite magnitude sense,

Fall 2013 MTH431/531 Real analysis Section Notes

Metric Space Properties

ACO Comprehensive Exam 9 October 2007 Student code A. 1. Graph Theory

Math 778S Spectral Graph Theory Handout #3: Eigenvalues of Adjacency Matrix

Complex Analysis Spring 2001 Homework I Solution

TMA4205 Numerical Linear Algebra. The Poisson problem in R 2 : diagonalization methods

1 lim. f(x) sin(nx)dx = 0. n sin(nx)dx

2 Banach spaces and Hilbert spaces

PROBLEM SET 5 SOLUTIONS 126 = , 37 = , 15 = , 7 = 7 1.

Slide Set 13 Linear Model with Endogenous Regressors and the GMM estimator

Continuous Functions

Lecture 8: October 20, Applications of SVD: least squares approximation

Chapter Vectors

Chapter 2. Periodic points of toral. automorphisms. 2.1 General introduction

CHAPTER 10 INFINITE SEQUENCES AND SERIES

Lecture 2. The Lovász Local Lemma

f n (x) f m (x) < ɛ/3 for all x A. By continuity of f n and f m we can find δ > 0 such that d(x, x 0 ) < δ implies that

MATH 413 FINAL EXAM. f(x) f(y) M x y. x + 1 n

Chapter 4. Fourier Series

Chapter 6 Principles of Data Reduction

Most text will write ordinary derivatives using either Leibniz notation 2 3. y + 5y= e and y y. xx tt t

On Involutions which Preserve Natural Filtration

MATH 205 HOMEWORK #2 OFFICIAL SOLUTION. (f + g)(x) = f(x) + g(x) = f( x) g( x) = (f + g)( x)

Axioms of Measure Theory

(3) If you replace row i of A by its sum with a multiple of another row, then the determinant is unchanged! Expand across the i th row:

Riemann Sums y = f (x)

1.3 Convergence Theorems of Fourier Series. k k k k. N N k 1. With this in mind, we state (without proof) the convergence of Fourier series.

Brief Review of Functions of Several Variables

Math 299 Supplement: Real Analysis Nov 2013

Estimation for Complete Data

Singular value decomposition. Mathématiques appliquées (MATH0504-1) B. Dewals, Ch. Geuzaine

Convergence of random variables. (telegram style notes) P.J.C. Spreij

MATH4822E FOURIER ANALYSIS AND ITS APPLICATIONS

A widely used display of protein shapes is based on the coordinates of the alpha carbons - - C α

If a subset E of R contains no open interval, is it of zero measure? For instance, is the set of irrationals in [0, 1] is of measure zero?

Resampling Methods. X (1/2), i.e., Pr (X i m) = 1/2. We order the data: X (1) X (2) X (n). Define the sample median: ( n.

Topologie. Musterlösungen

Recursive Algorithms. Recurrences. Recursive Algorithms Analysis

Weighted Approximation by Videnskii and Lupas Operators

For a 3 3 diagonal matrix we find. Thus e 1 is a eigenvector corresponding to eigenvalue λ = a 11. Thus matrix A has eigenvalues 2 and 3.

MATH 10550, EXAM 3 SOLUTIONS

10-701/ Machine Learning Mid-term Exam Solution

MA131 - Analysis 1. Workbook 2 Sequences I

Let us give one more example of MLE. Example 3. The uniform distribution U[0, θ] on the interval [0, θ] has p.d.f.

Sequences I. Chapter Introduction

Seunghee Ye Ma 8: Week 5 Oct 28

MAT1026 Calculus II Basic Convergence Tests for Series

Transcription:

Precoditioed fiite elemets method Let V be a Hilbert space, (, ) V a ier product o V ad V the correspodig iduced orm. Let a be a coercive, cotiuous, biliear form o V, that is, a : V V R ad there exist m, M, 0 < m M such that for all u, v, w V, α, β R, a(αu + βv, w) = αa(u, w) + βa(v, w), a(u, αv + βw) = αa(u, v) + βa(u, w) (biliearity), a(u, v) M u V v V (cotiuity), a(v, v) m v 2 V (coercivity). Let V be the set of all cotiuous, liear forms o V. The the followig Lax-Milgra results hold: (LM1) For ay F V, there exists a uique elemet u = u F a(u, v) = F(v), v V. V such that (LM2) If, moreover, V h is a fiite-dimesioal subspace of V, the to F we ca also associate a elemet u h V h such that a(u h, v h ) = F(v h ), v h V h, which is uiquely defied too. Approximatig u by u h Ituitively, such elemet u h ca be used as a approximatio of u if V h belogs to a family of subspaces {V h } h 0 of icreasig dimesio, such that the closure of h 0 V h coicides with V. I fact, it ca be show that a hypothesis of cosistecy o {V h } h 0 (implyig the latter property) yields the result: h 0 u u h V 0. (cov) Cosistecy of {V h } h 0 i V. {V h } h 0, V h V, is said to be cosistet i V if there exist V V dese i V (with respect to V ) ad a operator R h : V V h such that for ay v V, R h (v) v V 0 as h 0 (R h might be a iterpolatio operator). Let us show that the cosistecy of {V h } h 0 implies (cov). First we prove that the error u u h V is proportioal to the miimal error we ca have with V h. Note that a(u, v h ) = F (v h ), a(u h, v h ) = F (v h ) a(u u h, v h ) = 0 v h V h ad this remark implies m u u h 2 V a(u u h, u u h ) = a(u u h, u v h ) M u u h V u v h V, u u h V M m if u v h V. v h V h (cea) Now we ca prove (cov). Let v V be such that v u V < ε. By (cea) ad the cosistecy hypothesis, if h < h ε (h is suitably small), the u u h V M m u R h(v) V M m ( u v V + v R h (v) V ) < M m 2ε. 1

How to compute u h Let N h be the dimesio of V h, ad ϕ i, i = 1,..., N h, a basis for V h. The u h = N h (u h) j ϕ j ad the coditio a(u h, v h ) = F (v h ), v h V h, ca be rewritte as follows: N h (u h ) j a(ϕ j, ϕ i ) = F (ϕ i ), i = 1,..., N h. So the (u h ) j defiig u h ca be obtaied by solvig a liear system Ax = b, beig a ij = a(ϕ j, ϕ i ), b i = F (ϕ i ), 1 i, j N h. It is importat to otice that the symmetric part of the coefficiet matrix A is positive defiite, that is z T Az > 0, z R z 0. I fact, by the coercivity of a, we have z T Az = ij z i a(ϕ j, ϕ i )z j = a( z j ϕ j, z i ϕ i ) m z i ϕ i 2 V > 0 uless the z i are all ull (the ϕ i are assumed liearly idepedet). Example: a differetial problem solved by the fiite elemet method Assume that u : R is the uique solutio of the differetial problem (α u) + β u + γu = f, x, u = ϕ, x Γ D, u c = ψ, x Γ N. Here is a ope set i R d, Γ D ad Γ N are ope subsets of such that = Γ D Γ N, α : R d2, β : R d, γ, f : R. The, for all v, v ΓD = 0, a(u, v) := α u v + β uv + γuv = fv + ψvdσ. Γ N If we set u = u ϕ + w with u ϕ, w : R, u ϕ ΓD = ϕ ad w ΓD = 0, the the latter equatio becomes: a(w, v) = fv + ψvdσ a(u ϕ, v) =: F (v). Γ N So, we have the followig Problem. Fid w, w ΓD = 0 such that a(w, v) = F (v), v, v ΓD = 0. Moreover, the fuctios w, v must be also such that a(w, v) ad F (v) are well defied, that is we also require w, v H 1 () where H 1 () = {v L 2 () : D i v L 2 ()} (D i v L 2 () iff g i L 2 () g iϕ = vd iϕ ϕ C 0 () (D iv := g i )). Briefly, fid w H 1 0,Γ D () a(w, v) = F (v), v H 1 0,Γ D (). Uder suitable coditios o the data, α, β, γ, ϕ, ψ, the space V = H 1 0,Γ D () is a Hilbert space with respect to the ier product (u, v) V = (u, v) 1, = (u, v) L 2 () + i=1...d (D iu, D i v) L 2 (), ad the forms a ad F are well defied ad satisfy the coditios required by the Lax-Milgra results (LM1) ad (LM2) to hold. So, the w of the problem is well defied, ad for ay fiite-dimesioal 2

subspace V h of V = H 1 0,Γ D () is well defied a fuctio w h V h such that a(w h, v h ) = F (v h ), v h V h. Defiitio of w h coverget to w I order to yield fuctios w h coverget to w as h 0, we oly have to defie V h such that {V h } h 0 is cosistet i H 1 0,Γ D (). Let us do this i the case d = 2, = polygo, by usig the fiite elemet method. Let τ h be a triagulatio of of diameter h, that is a set of triagles T such that T τ h T = T ad diam (T ) =: h T h := max T τh h T, T τh T =, T 1, T 2 τ h T 1 T 2 is a commo vertex, a commo side, the whole triagle T 1 = T 2 or the empty set. To ay T i the followig we eed to associate also the umber ρ T, the diameter of the circle eclosed i T. Let S h be the set of all fuctios p : R such that p T is a degree-1 polyomial (i x 1 ad x 2 ) ad set Vh = S h C 0 (). Let i = 1, 2,..., Nh be the odes of the triagulatio τ h (the verteces of the triagles of τ h ) ad deote by ϕ i the elemets of Vh satisfyig the idetities ϕ i (j) = δ ij, i, j = 1, 2,..., Nh. Obviously ay elemet v of V h ca be expressed as v = N h i=1 v(i)ϕ i. Choose V h = Vh H1 0,Γ D () = Spa {ϕ 1,..., ϕ Nh } where 1,..., N h are the odes of the triagulatio τ h which are ot o Γ D. We wat to show that {V h } h 0 is cosistet i H0,Γ 1 D (), so that the well defied fuctios w h =...N h (w h ) j ϕ j V h such that a(w h, v h ) = F (v h ), v h V h, strogly coverge to w as h 0, i.e. w w h V 0. First we itroduce the space V = H0,Γ 1 D () H 2 (), cotaied ad dese i V = H0,Γ 1 D () (H 2 () = {v H 1 () : D ij v L 2 ()}, (u, v) 2, = (u, v) 1, + ij (D iju, D ij v) 0,, v 2 2, = i D iiv 2 0, ). Now let v be a elemet of such V. Notice that v C 0 () (...), thus the fuctio Π h v = N h i=1 v(i)ϕ i of V h, iterpolatig v i the odes of the triagulatio, is well defied. Moreover, Π h v is a fuctio of H 1 () ad oe ca measure the iterpolatig error usig the orm of V: v Π h v 1, c e h v 2,, c e costat. The latter iequality holds if the family of triagulatios {τ h } h 0 is chose regular, that is there exists a costat c r such that h T /ρ T c r for all T τ h ad h. Thus we have the operator R h : V V h required by the cosistecy hypothesis, it is the iterpolatig operator Π h. Observe that i case the fuctio w is i H 2 () we ca say more: w w h V 0 at least at the same rate of h. I fact, w w h V M m w Π hw V M m c eh w 2,. Computatio of w h I order to compute w h = N h (w h) j ϕ j oe has to solve the liear system Ax = b, a ij = a(ϕ j, ϕ i ), b i = F (ϕ i ). 3

I fact, (w h ) j = (A 1 b) j. More specifically, i our example, if s(g) deotes the set supp (g), the the etries of A ad b are: a ij = s(ϕ j) s(ϕ i) α ϕ j ϕ i + s(ϕ j) s(ϕ i) β ϕ jϕ i + s(ϕ j) s(ϕ i) γϕ jϕ i, b i = s(ϕ i) fϕ i + Γ N s(ϕ i) ψϕ idσ s(ϕ i) s(u ϕ) α u ϕ ϕ i s(ϕ i) s(u ϕ) β u ϕϕ i s(ϕ i) s(u ϕ) γu ϕϕ i, 1 i, j N h. Here the ϕ i are the Lagrage basis of V h (ϕ i (j) = δ ij ). So, the (w h ) j are the values of w h i the odes j ((w h ) j = w h (j)) ad the matrix A is sparse, i fact for ay fixed i, the umber of j such that the measure of s(ϕ j ) s(ϕ i ) is ot zero is smaller tha a costat (with respect to h) depedet upo the regularity parameter c r of the triagulatios (such costat is a boud for the umber of odes j liked directly to i). But these properties are far from to be essetial: i particular, more importat would be to kow that the matrix A is well coditioed. Ufortuately, eve i case the differetial problem is simply the Poisso problem (α = I, β = 0, γ = 0, Γ N = ) the matrix A has a coditio umber growig as (1/h) 2, if the Lagrage fuctios are used to represet w h. (This estimate of the coditio umber holds more i geeral for the covectio-diffusio problem, if the triagulatios are quasi-uiform (i.e. h T c u h, T τ h h) ad regular). So, cosider a arbitrary basis { ϕ i } of V h, ad represet w h i terms of this basis: w h = N h (w h ) j ϕ j. The, (w h ) j = (Ã 1 b) j where ã ij = s( ϕ j) s( ϕ i) ã ij = a( ϕ j, ϕ i ), bi = F ( ϕ i ), α ϕ j ϕ i + β ϕ j ϕ i + s( ϕ j) s( ϕ i) s( ϕ j) s( ϕ i) bi = s( ϕ i) f ϕ i + Γ N s( ϕ i) ψ ϕ idσ s( ϕ i) s(u ϕ) α u ϕ ϕ i s( ϕ i) s(u ϕ) β u ϕ ϕ i s( ϕ i) s(u ϕ) γu ϕ ϕ i. γ ϕ j ϕ i, If the ϕ i are such that µ 2 (Ã) < µ 2(A) (...), the we ca solve the system Ãx = b, better coditioed tha Ax = b, ad the, if eeded, recover w h = (w h (j)) N h = A 1 b solvig the system Sw h = w h = ((w h ) j ) N h = Ã 1 b. (Of course, all this is coveiet if S is a matrix of much lower complexity tha A). Let us prove this assertio i detail. Let v h be a geeric elemet of V h ad let S be the matrix such that ṽ h = Sv h, beig v h = [(v h ) 1 (v h ) Nh ] T ad ṽ h = [(v h ) 1 (v h ) Nh ] T such that v h V h N h N h (v h ) j ϕ j = v h = (v h ) j ϕ j. The we have ad therefore N h ϕ s = [S T ] sj ϕ j, s = 1,..., N h, a ij = a( N h r=1 [ST ] jr ϕ r, N h m=1 [ST ] im ϕ m ) = r,m [ST ] im a( ϕ r, ϕ m )[S] rj = r,m [ST ] im ã mr [S] rj = [S T ÃS] ij, 4

N h N h b i = F ( [S T ] ij ϕ j ) = [S T ] ij F ( ϕ j ) = [S T b]i. Thus, the equalities A = S T ÃS ad b = S T b must hold, ad the thesis follows. I the Poisso case, u = f, x, u = ϕ, x, a basis { ϕ i } for V h ca be itroduced yieldig a matrix à whose coditio umber µ 2(Ã) grows like (log 2 (1/h)) 2. (A aalogous result i the covectio-diffusio case (a ot symmetric) i 1995 was ot kow!). We ow see (ot i all details) that this is possible by usig a particular family of triagulatios τ h. Let τ 0 be a rada triagulatio of. Let us defie τ 1. For each triagle T of τ 0 draw the triagle whose verteces are the middle poits of the sides of T. The four triagles you see (similar to T ) are the triagles of τ 1. Note that if h 0 is the diameter of τ 0 (h 0 = max{h T : T τ 0 }), the h 1, the diameter of τ 1, is equal to 2 1 h 0. Note also that the odes of τ 0 are odes of τ 1 ; the ew odes of τ 1 are the middle poits of the sides of τ 0. We ca cotiue i this way, ad defie the triagulatios τ 2, τ 3,..., τ j,... (obviously, τ j is a abbreviatio for τ hj ). The diameter of the geeric τ j is h j = 2 j h 0, ad the family of triagulatios {τ j } + j=0 is regular ad quasi-uiform. To each τ j we ca associate the space V j = Vh j H0 1 () of the fuctios which are cotiuous, ull o, ad degree-1 polyomials i each T τ j. Note that V j V j+1. Let x jk, k I j = {1,..., N j } {1,..., Nj }, deote the geeric ier ode of the triagulatio τ j, ad {ϕ jk : k I j } the Lagrage basis of V j, ϕ jk (x jl ) = δ kl, k, l I j. Obviously, ay v V j ca be represeted as v = k I j v(x jk )ϕ jk, ad, if Π j is the iterpolatig operator, the v C 0 () Π j (v) = k I j v(x jk )ϕ jk. Istead of v(x jk ) we will write shortly v jk. Now that all is defied, cosider a fuctio v V j+1 ad observe that v j+1,k ϕ j+1,k = v = Π j+1 v = Π j v + (v Π j v) = v jk ϕ jk + (v Π j v). k I j+1 k I j Now the questio is: what must we add to V j i order to obtai V j+1? This questio ca be reduced to: what elemets of {ϕ j+1,k : k I j+1 } are eeded to represet v Π j v? Let x be a poit of ad let T be a triagle of τ j icludig x. Let us observe the above quatities ad, i particular, the fuctio v Π j v o T. Call x jk1, x jk2, x jk3 (k 1, k 2, k 3 I j ) the verteces of T. Note that they are odes also of τ j+1, thus x jki = x j+1,ρ(ki), for some ρ(k i ) Ij+1 o = {k I j+1 : x j+1,k is a ode of τ j }. Call x j+1,σ(k1k 2), σ(k 1 k 2 ) Ij+1 = I j+1\ij+1 o, the middle poit of the side x jk1 x jk2 of T, which is a ew ode, a ode of τ j+1, but ot of τ j. Draw the restrictios to T of the fuctios v ad Π j v. The it is clear that, o T, v Π j v = [v j+1,σ(k1k 2) 1 2 (v j,k 1 + v j,k2 )]ϕ j+1,σ(k1k 2) +[v j+1,σ(k2k 3) 1 2 (v j,k 2 + v j,k3 )]ϕ j+1,σ(k2k 3) +[v j+1,σ(k3k 1) 1 2 (v j,k 3 + v j,k1 )]ϕ j+1,σ(k3k 1) = ṽ j,σ(k1k 2)ϕ j+1,σ(k1k 2) + ṽ j,σ(k2k 3)ϕ j+1,σ(k2k 3) + ṽ j,σ(k3k 1)ϕ j+1,σ(k3k 1) = k I ṽ j+1 jk ϕ j+1,k 5

where ṽ jk = v j+1,k 1 2 (v j,k + v j,k ), k I j+1, ad k, k I j are such that x jk = x j+1,ρ(k ), x jk = x j+1,ρ(k ) are the extreme poits of the side of τ j havig x j+1,k as middle poit. Thus, if ψ jk := ϕ j+1,k = ϕ j+1,σ(k k ), k I j+1, the v V j+1 v = v j+1,k ϕ j+1,k = v jk ϕ jk + k I j+1 k I j k I j+1 ṽ jk ψ j,k. It follows that V j+1 = V j + W j, W j := Spa {ψ jk : k Ij+1 }, ad the set {ϕ jk : k I j } {ψ j,k : k Ij+1 } is a alterative basis of V j+1. So, the aswer to the questio is: the Lagragia fuctios of V j+1 correspodig to the ew odes. Observe that the v jk, k I j, ad the ṽ jk, k Ij+1, ca be computed from the v j+1,k, k I j+1, via the formulas v jk = v j+1,ρ(k), k I j ṽ j,k = v j+1,k 1 2 [v j+1,ρ(k ) + v j+1,ρ(k )], k Ij+1. Viceversa, the v j+1,k, k I j+1, ca be computed from the v jk, k I j, a from the ṽ jk, k Ij+1, via the formulas: v j+1,ρ(k) = v jk, k I j v j+1,k = ṽ j,k + 1 2 [v j,k + v j,k ], k I j+1. These formulas ca be writte i matrix form: v jk [ ] v j+1,ρ(k) k I j ṽ jk = I Ij 0 k I j B I I k Ij+1 j+1 v j+1,k k Ij+1, v j+1,ρ(k) k I j v j+1,k k I j+1 where each row of B has oly two ozero elemets, both equal to 1 2. = [ I Ij 0 B I I j+1 Now let J be a positive iteger. We are ready to itroduce a basis ϕ J,k, k I J, of the space V J yieldig a matrix Ã, ã r,s = a( ϕ J,s, ϕ J,r ), r, s I J, whose coditio umber i the Poisso case grows like O((log 2 (1/h J )) 2 ) = O((J + log 2 (1/h 0 )) 2 ), ad thus is smaller tha the coditio umber of the matrix A, a r,s = a(ϕ J,s, ϕ J,r ), r, s I J, yielded by the Lagrage basis ϕ J,k, k I J of V J (recall that µ 2 (A) = O((1/h J ) 2 ) = O((2 J /h 0 ) 2 )). Observe that if v V J the k I J v Jk ϕ Jk = v = Π J v = Π 0 v + J 1 j=0 (Π j+1v Π j v) = k I 0 v 0k ϕ 0k + J 1 j=0 ṽ jk ψ j,k, k I j+1 ψ jk = ϕ j+1,k, k I j+1, j = 0,..., J 1. It follows that V J admits the represetatio V J = V J 1 + W J 1 = V J 2 + W J 2 + W J 1 = V 0 + W 0 +... + W J 1 ] v jk k I j ṽ jk k I j+1 6

ad the set { ϕ J,k : k I J } := {ϕ 0k : k I 0 } {ψ 0,k : k I 1 } {ψ J 1,k : k I J } is a alterative basis of V J. Remark. Observe that if v J = (v J,k ) k IJ, ṽ J = (ṽ J,k ) k IJ = ((v 0,k ) k I0 (ṽ 0,k ) k I 1 (ṽ J 1,k ) k I J ), the ṽ J = Sv J = E 0 P 0 E 1 P 1 E J 1 P J 1 v J where the P k are permutatio matrices, the E k are matrices of the form I I0 0 I I1 0 [ E 0 = B 0 I I 1, E 1 = B 1 I I I IJ 1 0 2,..., E J 1 = B J 1 I I I I J ] ad the B k, i the defiitio of the E k, have oly two ozero elemets for each row, both equal to 1 2. So, v J ca be computed from ṽ J (as well as ṽ J ca be computed from v J ) with 2( I J I 0 ) divisios by 2. The trasform of v J ito ṽ J is described i detail here below: v J = [ vj,k k I J ] P J 2 E J 1 P J 1 v J = P J 1 v J = v J 1,ρ(k) k I J 2 v J 1,k k I J 1 ṽ J 1,k k I J v J,ρ(k) k I J 1 v J,k k IJ E 0 P 0 E J 1 P J 1 v J = E J 1P J 1 v J = E J 2 P J 2 E J 1 P J 1 v J = v 0,k k I 0 ṽ 0,k k I 1 ṽ J 1,k k I J. v J 1,k k I J 1 ṽ J 1,k k I J Theorem. Let τ j, V j, Π j, j = 0, 1,..., J, be the triagulatios of, the subspaces of V = H0 1 () ad the iterpolatig operators C 0 () V j defied above. For ay v V J set J 1 ˆ vˆ 2 = Π 0 v 2 1, + j=0 k Ij+1 J 1 (Π j+1 v Π j v)(x j+1,k ) 2 = Π 0 v 2 1, + j=0 k Ij+1 v J 2,k k I J 2 ṽ J 2,k k I J 1 ṽ J 1,k k I J ṽ j,k 2. 7

The there exist two positive costats c 1, c 2 (depedig oly o the agles of τ 0 ) such that c 1 ˆ vˆ 2 J 2 v 2 1, c 2ˆ vˆ 2. This iequality, ivolvig the coefficiets ṽ J,k of v V J with respect to the hierarchical basis ϕ J,k, k I J, is due to Yseretat. It allows us to evaluate the coditio umber of Ã, ã r,s = a( ϕ J,s, ϕ J,r ), r, s I J, i the Poisso case where a(u, v) = u v. First ote that Π 0 v 2 1, = k I 0 v 0,k ϕ 0,k 2 1, = ( k I 0 v 0,k ϕ 0,k ) ( k I 0 v 0,k ϕ 0,k ) = k,si 0 v 0k v 0s ϕ 0k ϕ 0s, thus the Yseretat iequality ca be rewritte as follows: [ ] N 0 ṽj T ṽ 0 I J [ c 1 ṽj T Mṽ J c 2 ṽj T N 0 0 I J 2 ] ṽ J where N = ( ϕ 0r ϕ 0,s ) r,s I0 ad M = ( ϕ J,r ϕ J,s ) r,s IJ. Note that N ad M are positive defiite matrices. Note also that i case of the Poisso differetial problem u = f, x, u = ϕ, x, the form a is simply a(u, v) = u v, i.e. we have the cotiuous problem w V = H0,Γ 1 D () w v = fv u ϕ v, v V = H0,Γ 1 D () which is reduced first to the discrete problem w J V J w J v J = fv J ad the, via the represetatio w J = k I J u ϕ v J, v J V J (Y) (w J ) k ϕ J,k, to the liear system Ãx = b, ã r,s = ϕ J,r ϕ J,s, br = f ϕ J,r u ϕ ϕ J,r, (w J ) k = (à 1 b)k. Observe that the coefficiet matrix à of this system is exactly the matrix M i 1 (Y). Now we prove that µ 2 (M) = O((log 2 h J ) 2 ). Cosider the Cholesky factorizatio of N, N = L N L T N, ad ote that [ ] [ ] LN N L := LL I T =. I Set z = L T ṽ J, ṽ J = L T z. By (Y), for all vectors z 0 we have 1 c 1 J 2 zt L 1 ML T z z T c 2. z Thus, if λ is ay eigevalue of the positive defiite matrix L 1 ML T, the c 1 J 2 λ c 2, ad this result implies that the coditio umber of L 1 ML T is bouded by c 2 c 1 J 2. Sice L N is a small matrix ad its dimesio does ot deped o J, it follows that µ 2 (M) cj 2 = c(log 2 h 0 h J ) 2 for some costat c. 8