Technical Guide to Upgrading/Migrating to XenApp 5 Feature Pack 2

Technical Guide to Upgrading/Migrating to XenApp 5 Feature Pack 2

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Technical Guide to Upgrading/Migrating to XenApp™ 5 Feature Pack 2 Citrix released XenApp 5 Feature Pack 2 on September 29, 2009. This document provides a brief technical overview of the feature set to help you determine which features are applicable to your environment and then focuses on easy-to-follow installation steps. It should serve as a technical guide for Citrix administrators and partners. Author: Jo Harder, Citrix Worldwide Technical Readiness Date Created: October 2009
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13: BrezziFortinstability of saddle point problems. Math 639 (updated: January2, 2012)
In this section, we consider a saddle point problem.This is defined from a matrix (with real entries) given in the following block form:   t A B A=. B0 t HereAhas dimensionn×n,Bhas dimensionm×nandB(the transpose ofB) has dimensionn×mGiven. Weseek to iteratively solve the problem: n+m n+m bR, findxRsolving (13.1)Ax=b. This is the simplest of saddle point problems.The techniques this section as well as their application to more general saddle point problems can be found in BrezziFortin, ... Remark 1.We temporarily consider the case whenAis symmetric and positive definite and the rank ofBism. Theminimax characterization of the eigenvalues ofA, i.e., (Az, z) λj= max min. SjzSj(z, z) Here the minimum is taken over all subspacesSjof dimensionjandλ1λ2...are the eigenvalues ofAin decreasing order.The positive definite ness ofAimmediately implies thatAhas at leastnpositive eigenvalues since n forjn, takingSj= [V,0]whereVis ajdimensional subspace ofRgives rise to a positive minimum. Similarly, the eigenvalues ofA, given in increasing order are given by (Az, z) µj= minmax. SjzSj(z, z) m1t1t t ForqR, we setz= [A Bq, q]and find that(Az, z) =(qA Bq, B). t Since the rank ofB=m, it follows thatAhas at leastmnegative eigenval ues, i.e.,Ahasnpositive andmnegative eigenvalues and(13.1)is a saddle point problem.
We shall recast this problem in a slightly different notation by defin t t ingu= (x1, . . . , xn) ,p= (xn+1, . . . , xn+m)F= (b1, . . . , bn) andG= 1
2 (bn+1, . . . , bn+m(). Then13.1) is equivalent to the system of equations t Au+B p=F (13.2) Bu=G. n+m As we shall often be splitting vectors inR, we introduce the notation x= [u, p] andb= [F, G] for this splitting. Before investigating the iterative solution of (13.1), we consider some conditions which guarantee the existence of solutions to (13.1) for any vector n+m1 bR n m We let (x, y) =xydenote the Euclidean inner product onR,Ror n+m Rdepending on where the vectorsxandyreside. Wealso introduce n m auxiliary norms onRandRwhich we denote bykxk(again, the specific norm is identified from wherexnorms are induced by innerresides). These n mn+m products onRandR, respectively.Forz= [w, q]R, we set p 2 2 kzk=kwk+kqk. We define the following (dual) norms (Ax, y) kAk= supsup, kxkkyk n n xRyR (Bx, y) kBk= supsup. kxkkyk n m xRyR It is immediate that t (B y,x) t kBksup= sup kxkkyk m n yRxR (Bx, y = supsup =kBk kxkkyk m n yRxR where we reversed the order of the supremums to derive the last equality above. n We shall use additional dual norms for vectors, i.e., forwR, (w, x) kwk= sup. kxk n xR m The dual normskqkandkykforqRandyRn+mare defined analogously. Itis not hard to show that the dual norms and the matrix
1 Of course,Ais a square matrix and so existence immediately implies existence.
3 norms do indeed satisfy the norm axioms.There are obvious inequalities which are immediate from the definition of these norms, e.g., |(p, q)| ≤ kpkkqkandkAuk≤ kAk kuk. n+m Exercise 1.For allz= [w, q]R, kzk≤ kwk+kqk2kzk. Conditions on the matrixA: (M.1) ThematricesAis symmetric and positive semidefinite. (M.2) Wedenote the null space ofBby n kerB={xR:Bx=0} and assume thatAis coercive on kerBspecifically, we assume. More that there is a positive constantαsatisfying 2 αkxk ≤(Ax,x),for allxkerB. (M.3) We assume the following infsup condition, i.e., there is a positive constantβsatisfying (Bu, p) (13.3)βinf sup. m uRkukkpk pRn Remark 2.We note that(13.3)implies that (Bu, p) m (13.4)kpk ≤csupfor allpR kuk n uR with constantc= 1fact,. Inc= 1is the smallest constantcfor which (13.4)holds. These systems often come from the discretization of partial differential equations with constraints enforced by the introduction of Lagrange mul tipliers (discussed below).In this case, natural problem dependent norms n m (which we have denotedkukforuRandkpkforpR) often give rise to 11 constantsα,β,kAk,kBkwhich are uniformly bounded independently of the discretization or mesh size.These norms are central to the stability analysis of the original PDE as well as their discretizations (i.e., the solution of (13.1)). It is a fact from linear algebra that n t R= kerBRangeB
4 is an orthogonal decomposition in the Euclidean inner product and we denote t m kerB= RangeB(. Now,13.4) implies that for each nonzeropR, there n is at least one (nonzero)uRwith t (Bu, p) = (u, Bp)6= 0. t mThis means thatBis a one to one map ofRonto kerBit follow. Thus t tfrom the (13.4) thatB(the inverse ofBon kerB) satisfies t tt (uBw, B) (B uw, B) t11 kB uk ≤βsup =βsup kwk kwk n n wRwR (13.5) (w, u) 11 =βsup =βkuk. kwk n wR Now there is a complementary infsup condition, namely, (Bu, p) (13.6)β1inf sup. kukkpk t m ukerB pR or (Bu, p) 1t (13.7)kuk ≤βsup forallukerB . 1 pRkuk m It turns out that (13.3) and (13.6) are equivalent and hold with the sameβ. We illustrate the proof in one direction below. tt Suppose that (13.3) holds and thatuis in kerBfor. Thenp=B u, (13.5) implies t (u, u) (pu, B) kuk= = kuk kuk (Bu, p) (Bu, p) 1 =βsup kukpRkpk m 11 This implies that (13.6) holds withββ. Asimilar argument shows 1 11 that if (13.6) holds then so does (13.3) withββ, i.e., the infsup 1 conditions are equivalent and hold with the same constant. m1 NowBis one to one on kerBand maps ontoR. ItsinverseBmaps mRbijectively onto kerBinfsup condition (. The13.7) implies 1 (qBB p,) (p, q) 111 kB pk ≤βsup =βsup kqk kqk m m qRqR (13.8) (w, u) 11 =βsup =βkuk. wRkwk n
5 n+m n+m Theorem 1.Letbbe inRthere is a unique solution. ThenxR satisfying(13.1). Moreover,there is a constantc=c(α, β,kAk,kBk)such that kxk ≤ckbk. Proof.We shall construct the solution [u, pof (13.2). Wefirst defineu0= 11 B G. By(??),ku0k ≤βkGk. Next,we setu1kerBto be the solution of (13.9) (Au1, θ) = (F, θ)(Au0, θall) forθkerB. That this problem has a unique solution follows from the Riesz Represen tation Theorem applied on kerBwith inner product< u,v >= (Au, v), u, vkerB. That<,>is an inner product on kerBfollows from the coercivity assumption (M.2).We can then boundu1by 2 αku1k ≤(Au1, u1) = (F, u1)(Au0, u1) ≤ kFkku1k+kAk ku0kku1k which immediately implies 11 (13.10)ku1| ≤α(kFk+βkGk). We setu=u0+u1and note that 11 (13.11)kuk ≤(1 +α)β1kGk+αkFkSet t (13.12)p=B(FAu). Note that (13.9) implies thatFAuis in kerBand hencepis well defined and satisfies (13.13) 11 kpk ≤ kFk+kAk kuk ≤ kFk+kAk((1 +α)β1kGk+αkFk). Note that (13.12) implies the first equation in (13.2) while Bu=Bu0+Bu1=Bu0=G. This implies that [u, p] is the solution of (13.2bound in the theorem). The follows from (13.11), (13.13) and Exercise1.Remark 3.The assumption of symmetry onAcan be relaxed by replacing the Riesz Representation Theorem by the LaxMilgram Theorem. Corollary 1.There are constantsc0andc1depending only onα,β,kAk, andkBksatisfying n+m (13.14)c0kxk ≤ kAxkc1kxkfor allxR.
6 n+m Proof.For anyxR,xis the unique solution to (13.1) withb=Ax thus the first inequality of (13.14) is the inequality of the theorem. For the second, letx= [u, pExercise]. By1, t kAxk≤ kAu+B pk+kBuk(kAk+kBk)kuk+kBkkpk (kAk+kBk)(kuk+kpk)2(kAk+kBk)kxk. 2 22 We used the arithmetic geometric mean inequality (a+b)2(a+b) for the last inequality above.