Math 5311 Problem Set #5 Solutions

Transcription

1 Math 5311 Problem Set #5 Solutios March 9, 009 Problem 1 O&S Part (a) Solve with boudary coditios u = 1 0 x < L/ 1 L/ < x L u (0) = u (L) = 0. Let s refer to [0, L/) as regio 1 ad (L/, L] as regio. I regio 1, the solutio is u 1 (x) = 1 x + A 1 x + B 1. I regio, the solutio is u (x) = 1 x + A x + B. At the iterface, we must have cotiuity of u ad u (see otes o iterface coditios) so that 1 L 4 + A L 1 + B 1 = 1 L 4 + A L + B L + A 1 = L + A. The boudary coditios gives us the equatios A 1 = 0 L + A = 0 = A = L. Notice that the secod iterface coditio is automatically satisfied by A 1 = 0, A = L, so that it is ot idepedet of the BCs. The system of equatios is therefore sigular, so we ll expect to be uable to solve for all coefficiets. Substitutio of the resultig A 1 ad A ito the first iterface coditio gives L /4 = B 1 B. We have o more equatios to work with, so we ca t determie B 1 or B other tha to say that their differece is L /4. Let s itroduce a ew costat B ad write B 1 = B + L /4 ad B = B. Notice that had the BCs ivolved the value of u rather tha u aloe, oe or both of B 1, B would have survived to appear i the BC equatios makig it possible to solve for them. With Neuma BCs, the best we ca do is specify the solutio to withi a additive costat B. The solutio is therefore u(x) = 1 x + L 4 + B 0 x < L 1 x + Lx + B L < x L. 1

2 Part (b) With the costat c = 1 the stress is equal to the strai, u x 0 x < L/ (x) = x + L L/ < x L which is idepedet of the choice of B. Part (c) Ay value of B ca be used, we ll use B = 0. Displacemet Strai Figure 1: Displacemet ad strai for B = 0 cofiguratio. Problem 1. By OS Thm 8.0 the eigevectors of K form a orthoormal basis for R. Cosequetly, we ca represet ay vector x R as x = x j v j. j=1 Now compute p(x) = x T Kx, By hypothesis, each λ i λ 1, so p(x) = = i=1 j=1 i=1 j=1 = x i x j (v T i Kv j x i x j vi, λ j v j xi λ i. i=1 ) p(x) λ 1 i=1 x i = λ 1

3 where i the last step the costrait x T x = 1 was used. Therefore, we have bouded p(x) from below by λ 1. To establish that λ 1 is the miimum of p, we also eed to show that x such that x = 1 ad p(x ) That step is easy: let x be the eigevector v 1, i which case x 1 = 1, x i =1 = 0 ad p(v 1 ) Therefore λ 1 is the miimum value of p(x) o the surface x = 1.. The Rayleigh quotiet is essetially a ormalized objective fuctio. (a) Represetig x i eigevectors, R(x) = i=1 x i λ i i=1. x i From part 1, we kow the umerator is λ 1 i=1 x i, so R(x) λ 1 i=1 x i i=1 x i (b) A straightforward calculatio establishes sufficiecy: R(αv 1 ) = α v T 1 Kv 1 α v T 1 v 1 To prove ecessity, we ll write x = αv 1 + j= ɛ jv j ad show that all ɛ j are ecessarily zero if R(x) = λ 1. With that represetatio of x, compute the Rayleigh quotiet Assume R(x) = λ 1, i which case R(x) = α λ 1 + j= ɛ j λ j α +. j= ɛ j λ 1 (α + j= ɛ j ) = λ 1α + j= ɛ j λ j. Itroduce µ j = λ j λ 1, ad subtract the left had side from the right had side, leavig 0 = µ j ɛ j. j= However, by the hypothesis that λ 1 has multiplicity 1 all µ j are strictly positive, so that the oly possible way the RHS ca be zero is for every ɛ j = 0. Thus, x = αv 1 is ecessary for R(x) = λ 1, ad we re doe. i. If there is some k = 1 such that λ k = λ 1, the proof i part (b) fails because whe µ k = 0 it becomes possible for ɛ k = 0. The Rayleigh quotiet i such a case becomes R(x) = α λ 1 + ɛ k λ k α + ɛ k = λ 1, providig a couterexample. Let s exted the theorem established i part (b) to the case whe λ 1 has multiplicity m 1. Let X R be the subspace spaed by the m eigevectors havig eigevalues equal to λ 1. Ay x X ca be writte as m x = α j v j j=1 ad the Rayleigh quotiet for x X is R(x) = m j=1 α j λ j m j=1 α j 3

4 Thus x X is sufficiet for R(x) The proof of ecessity is similar to that i part (b). We write x = m j=1 α jλ 1 + j=m+1 ɛ jλ j ad the show that ɛ j = 0 is icosistet with R(x) The Rayleigh quotiet is ad the assumptio that R(x) = λ 1 leads to Defiig µ j = λ j λ 1 > 0 as before gives R(x) = m j=1 λ 1α j + j=m+1 λ jɛ j m j=1 α j + j=m+1 ɛ j ɛ j λ 1 = ɛ j λ j. j=m+1 j=m+1 ɛ j µ j = 0. j=m+1 Therefore, ɛ j = 0, establishig that x X is ecessary for R(x) Notice the differece betwee part 1 ad part. I part 1, we restricted ourselves to the costrait surface x = 1, i part we made o such restrictio, but accouted for deviatios of x from 1 through the ormalizatio factor i the deomiator of the Rayleigh quotiet. I both cases, we ca compute the smallest eigevector by solvig a miimizatio problem; the tradeoff betwee the two is that miimizig the Rayleigh quotiet elimiates the difficulty of workig with the costrait, but at a price: R(x) is a more complicated objective fuctio tha p(x). This sort of tradeoff betwee complexity of objective fuctio ad complexity of costrait is commo i formulatig optimizatio problems. Problem 3. To fid the statioary poits of x T Kx subject to x T x = 1, set up the Lagragia L = x T Kx λ(x T x 1) ad fid its statioary poits. The Gateaux differetial of the Lagragia is At a statioary poit d (h,µ) L = 0 (h, µ), so d (h,µ) L = h T (Kx λx) µ T (x T x 1). Kx = λx x T x = 1. I other words, the statioary poits are the ormalized eigevectors of K. Whe the uormalized eigevectors are K = [ ] [ 1 + ] T [, 1 ad 1 T., 1] Normalizatio is straightforward. The geometry of the eigevectors is illustrated i Figure. The eigevectors poit towards the major ad mior axes of the elliptical cotours of the quadratic form. The Mathematica otebook used to draw this figure is available o the course web page. If you have a hard time visualizig how the various compoets of the problem fit together, rotatig the picture to view from several agles may help. 4

5 Figure : Illustratio of the geometry of miimizatio of a quadratic form p(x) = x T Kx i R subject to the costrait x T x = 1. The quadratic form is plotted i 3D as a curved surface z = p(x). Level curves of the quadratic form are show as cotour lies. The costrait x T x = 1 is show as the circular cylider. The itersectio betwee the costrait surface ad the quadratic form is show with the thick red curve. The blue lies i the (x, y) plae show the eigevectors; it is clear that the eigevectors poit to the positios of the miimum ad maximum values of p(x) alog the costrait curve. The blue ad red circles idicate the positios of the miimum ad maximum, respectively. As you proved i problem of this problem set, the fuctio values at these poits are just the eigevalues associated with the eigevectors defiig their positios i the (x, y) plae! Thus, the z coordiates of the blue ad red circles are the miimum ad maximum eigevalues, respectively. 5