Isem-comment-Nov09-1-daniel

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1Dear ISEM team,We have a few comments regarding lectures one and two. It is ouropinion that Lemmas 1.2, 2.1, and 2.2 could be presented in a more con-structive way. We therefore propose the following formulation which (1)gives more information about the transformation Q mentioned in Lemmad d 02.2 of lecture two and (2) shows how the spaceR and its dual space (R )are isomorphic to each other using the expression for Q and the transfor-mation J as given in Lemma 1.2 of lecture one and Lemma 2.1 of lecturetwo respectively. One of the motivations for this formulation is to introducea numerical aspect among the topics in this course. As is known, the roleof gradients plays an important part in numerical analysis and in particu-lar optimization problems (see for example the work John W. Neuberger,chapter 8 in Sobolev Gradients and Differential Equations, Lecture Notesin Mathematics 1670, Springer-Verlag (1997)). We thus suggest an exercisethat was taken from a course given by John Neuberger to demonstrate therelationship between gradient systems in an infinite dimensional Hilbertspace setting and finite dimensional spaces.We first propose the following Lemma 2.2 , which gives you more in-formation about the operator Q mentioned in the original Lemma 2.2 oflecture two: dLemma 2.2 For every inner producth ; i onR there exists an iso-d dmorphism Q :R ! R having the following propertiesd(i)hv;wi =hQv;wi for all v;w2R ,euc(ii) Q is symmetric,(iii) Q is ...

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• Publié par: Uffe
• Nombre de lectures: 12
• Langue: English

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Dear ISEM team,

1

We have a few comments regarding lectures one and two.It is our opinion that Lemmas 1.2, 2.1, and 2.2 could be presented in a more con- structive way.We therefore propose the following formulation which (1) gives more information about the transformationQmentioned in Lemma d d0 2.2 of lecture two and (2) shows how the spaceRand its dual space(R) are isomorphic to each other using the expression forQand the transfor- mationJas given in Lemma 1.2 of lecture one and Lemma 2.1 of lecture two respectively.One of the motivations for this formulation is to introduce a numerical aspect among the topics in this course.As is known, the role of gradients plays an important part in numerical analysis and in particu- lar optimization problems (see for example the work John W. Neuberger, chapter 8 in Sobolev Gradients and Differential Equations, Lecture Notes in Mathematics 1670, Springer-Verlag (1997)).We thus suggest an exercise that was taken from a course given by John Neuberger to demonstrate the relationship between gradient systems in an infinite dimensional Hilbert space setting and finite dimensional spaces. ∗ We first propose the following Lemma 2.2 , which gives you more in- formation about the operatorQmentioned in the original Lemma 2.2 of lecture two: ∗d Lemma 2.2For every inner producth∙,∙ionRthere exists an iso- d d morphismQ:R−→Rhaving the following properties d (i)hv, wi=hQv, wi euc for allv, w∈R, (ii)Qis symmetric, (iii)Qis positive, (iv)Qdefinite, d×d (v) thereexists a symmetric, positive-definite matrixA∈Rdependent d on the choice of the basis{e 1 , . . . , e d }ofRand the inner producth∙,∙i d such thatQu=Aufor allu∈R. d Proof. We take a basis{e 1 , . . . , e d }ofRand define the matrixA= d d×d (∈R a i,j ) i,j=1 bya i,j =he i , e j ifor alli, j= 1, . . . , d. Bythe definition of t the given inner producth∙,∙i, we have thatAis symmetric (i.e.,A=A) and t d positive-definite (i.e.,v Av >0for allv∈R\{0}define the mapping). We d dd Q:R−→RbyQu=Aufor allu∈Rand we see thatQis symmetric, positive-definite asAis. Sinceeach positive-definite matrix is invertible, −1−1d the inverse ofQis given byQ u=A ufor allu∈Rsince. Moreover, P d d allv, w∈Rcan be written as a linear combinationv=v i e i with i=1 P d unique scalarsv 1 , . . . , v d ∈R(w.r.t.{e 1 , . . . , e d }) andw=w j e j with j=1

2

w 1 , . . . , w d ∈R, we obtain  ! d d X X hv, wi=v i w j he i , e j i(by bilinearity ofh∙,∙i) i=1j=1 t =v Aw(by multipl.of matrices and def.ofA) =hv, Qwi euc (by def.ofQ) =hQv, wi euc (by symmetry ofQ)

∗ With Lemma 2.2we prove the Representation Lemma (Lemma 2.1) of lecture one: d d0 Proof. We define the linear operatorJ:R−→(R)by d hJ u, vi (R),R :=hu, vifor allu, v∈R, d0d d whereh∙,∙idenotes the given inner product onRCauchy-Schwarz’s. By d0 inequality, we find thatJ ubelongs to(R)and thatJUsingis bounded. the positive-definiteness ofh∙,∙igives thatJis injective.It remains to verify 0d0 the surjectivity ofJsee this we take some. Tou∈(R)and set 0t d v i =hu , e i i (R),R for alli= 1, . . . , d,v= (v 1 , . . . , v d )∈R, d0d d d∗ where{e 1 , . . . , e d }ofRis some choosen basis ofRLemma 2.2. Bythere d d is a symmetric positive-definite isomorphismQ:R−→Rsuch that d hu, wi=hQu, wi euc for allu, w∈R. (1) P d −1d d Thenu=Q vbelongs toRsuch that for allw=w i e i ∈Rwith i=1 w 1 , . . . , w d ∈R, hJ u, wi (R),R =hu, wiof(by def.J) d0d =hQu, wi euc (by equation (??)) =hv, wi euc d X =v i w i i=1 d X 0 0 =hu ,w i e i i (R),R (by linearity ofu) d0d i=1 0 =hu , wi (R),R . d0d ThereforeJis an isomorphism and hence Lemme 2.1 is proved.

3

Excercise: 1 - Motivation:Consider the infinite dimensional Hilbert spacesH(I) 2 andL(I)whereI⊂Ris a bounded interval.Then there is a linear 1 2 transformationMso that for allu∈H(I)andv∈L(I), one has hu, vi L(I) =hu, M vi H(I) . 2 1 −1 It is well-known thatM= (I−Δ N )whereΔ N is the Neumann Laplacian. - Now, we search a finite dimensional formulation for this problem. Thus we take a uniforme partitionP={a=t 0 <∙ ∙ ∙< t d =b}of the b−a intervalI= [a, b]with fixed time stepτ=>0(i.et i =a+iτfor d i= 0, . . . , d) and for every functionf:I−→Rwe set f i =f(t i )fori= 1, . . . , d, and define by t d+1 ~ f= (f 0 , f 1 . . . , f d )∈R thefinite dimensional versionoff. Thenwe denote by f i −f i−1 ~ (Df) i = (i= 1, . . . , d) τ th thefirst order finite differencingforfat theigrid point, and ~ Dfis given by   t d ~ ~~ Df= (Df) 1 , . . . ,(Df) d ∈R d If we denote byh∙,∙i euc,(d) the Euclidean inner product onR, then we d+1 define a new inner producth∙,∙ionRby

~ ~~ hg~,fi=hf,~gi euc,(d+1) +hDfD,g~i euc,(d)

d+1 ~ for all,~fg∈R.

∗ - Theexcercise is to find an expression forQas defined in Lemma 2.2 for this case.

Thank you for your time, Daniel Hauer and Parimah Kazemi