Math 273b: Calcls of Variations Yacob Kreh Homework #3 [1] Consier the 1D length fnctional minimization problem min F 1 1 L, or min 1 + 2, for twice ifferentiable fnctions : [, 1] R with bonary conitions, 1 1. a Show that the fnctional F is conve. b Formally compte the Gatea-ifferential an then obtain the Eler Lagrange eqation associate with the minimization. c Fin the eact soltion of the problem. Soltion: a To prove conveity of F, first prove it for L. Start by ifferentiating L twice. We get 2 1 + 2 1 2 2 1 + 2 1 2 2 1 + 2 1 2 1 + 2 1 1 2 + 2 1 + 2 3 2 2 1 + 2 1 2 2 1 + 2 3 2 1 + 2 1 2 1 2 1 + 2 1 1 + 2 1 2 1 2 1 + 2 1 + 2 1 1 2 >. 1 + 2 1 1
Yacob Kreh 2 This proves that L is conve an so, F λ + 1 λv 1 Lλ + 1 λv 1 λl + 1 λlv λf + 1 λf v. So F is conve. b To compte the Gatea erivative, consier an arbitrary smooth v sch that v v1 an an arbitrary ɛ >. We see we get F + ɛv F lim ɛ ɛ lim ɛ lim ɛ lim 1 1 1 ɛ 1 v 1 1 + + ɛv 2 1 1 + 2 ɛ 1 + + ɛv 2 1 + 2 ɛ 1 + + ɛv 2 1 + 2 v v 1 + 2 1 As v vanishes at bonaries, we are left with 1 ɛv 1 + 2 v. 1+ 2 minimizing, the Gatea erivative mst vanish for all v, that is 1 for all fnctions v which implies 1+ 2 3 2 1+ 2 2 v. Since we are v 1+ 2. Differentiating leas to so we get. This is the Eler Lagrange Eqation for this problem. c Since the secon erivative vanishes, the fnction mst be linear, an since only one line passes throgh, an 1,1, we get that the niqe soltion to the 1D length minimization is. [2] Let A : R n R n be a linear self-ajoint operator, b R n, an consier the qaratic fnction for R n Show that the three statements i inf{q : R n } > ii A O an b ImA. q : A, 2 b,.
Yacob Kreh 3 iii the problem inf{q : R n } > has a soltion are eqivalent. When they hol, characterize the set of minimm points of q, in terms of the pseo-inverse of A. Soltion: i ii We o this by contrapositive, sppose that A is not positive semi-efinite, then A has a negative eigenvale c < with some fie eigenvector, i.e. A c. Consier α α. Now look at q α A α, α 2 b, α T αa α 2b T α c α 2 2b T α cα 2 2 αb T. As c 2 an b T are fie, call these 1 < an 2, we can make q α 1 α 2 + 2 α as negative as we want as it is qaratic in α with negative leaing coefficient, ths inf{q}. This proves inf{q} > implies A. Net assme b ImA. Consier an orthonormal basis for A {v i } with eigenvales { }. We have that there oes not eist a soltion to b i v i b A A i v i i v i which means j sch that b j bt λ j. Let α αv j. Now we have q α A α, α 2 b, α, α 2 b i v i, α 2αb j, this can be mae as positive or negative as we want as it is linear in α. Ths again inf{q} >. This proves inf{q} > implies b ImA. ii iii As A we have that there eists an orthonormal eigenbasis {v i } whose eigenvales are all non-negative. Let s arrange the eigenvales in ecreasing orer repeate eigenvales are fine, any orer is okay an so we have that the first k eigenvales are positive an the last n k are, where n k is the imension of the kernel. We also have that b b i v i ImA an so we know that b i for i > k. Evalating q we get A, 2 b, A i v i, i v i 2 b i v i, i v i λi i 2 2 b i i i 2 2 b i i. Notice that we only nee to sm p to k as for i > k, b i. Ths we have that q k i 2 2b i i. This is jst a sm of qaratic polynomial in k variables with positive leaing coefficients, an it can be minimize by minimizing each separately as can be seen partial ifferentiation techniqes. We fin that setting i b i for i k is the minimizer, with i free in the other coorinates as they will vanish. Ths inf{q} > has a soltion. iii i Trivial. To constrct the pseo-inverse of A, let U be the n k matri whose colmns are the eigenvectors with nonzero eigenvales an D the iagonal k k matri consisting of those nonzero eigenvales. The pseo-inverse is then UD 1 U T. Or soltion set is UD 1 U T b + KerA. We note that U T b projects b onto to basis of the first k eigenvectors, which is jst a basis transformation since b is in the image of A so the projection oes not lose any information, we jst get the b i terms from the epansion b k b i v i. Then mltiplication by D 1 precisely gives b i an mtliplication by U gives s k b i v i. The pseo-inverse gives the soltion from the soltion set that minimizes, becase it oes not a anything from the kernel.
Yacob Kreh 4 [3] Comptation of the Eler-Lagrange eqation. a Consier the minimization problem inf F L,,, with, 1 1 given constants, an L a sfficiently smooth fnction. Obtain formally the Eler-Lagrange eqation of the minimization problem that is satisfie by a smooth optimal. Hint: Consier smooth test fnctions v, sch that v v 1. Since is a minimizer, we mst have F F + εv for all sch sfficiently smooth fnctions v an every real ɛ. Apply integration by parts to obtain the esire reslt. Yo shol obtain a secon-orer ifferential eqation. b Let now, t be a smooth soltion of the time-epenent partial ifferential eqation PDE t L,, L,,, with, on, 1 an, t U, 1, t U 1 for t. Show that the fnction Et F, t is ecreasing in time, where F 1 L,,. Soltion: a We want to minimize F, that is, if is the minimizer F F + w for all fnctions w sch that w vanishes at the bonary. So let s assme is sch a minimizer of F, now consier a smooth test fnction v sch that v v 1 an an ɛ >. Let s eamine Gɛ 1 L, +ɛv, +ɛv. We shol have that this Gɛ attains a minimm at ɛ, which we prove by taking erivative. Gɛ ɛ ɛ L, + ɛv, + ɛv where we have a total erivative. Let Lɛ L, + ɛv, + ɛv an L L,, an we get ɛ Lɛ Lɛ + ɛv Lɛ + ɛ ɛ + ɛv + + ɛv Lɛ ɛ + ɛv v Lɛ + ɛv + Lɛ v. + ɛv Now we can set ɛ an this integral shol vanish. That is, we get v L L + v. 3
Yacob Kreh 5 An integration by parts gives L v L v L 1. However, we know that v vanishes at the bonary, an since v is an arbitrary smooth fnction, L L. That is to say, the minimizer of satisfies that above secon orer ifferential eqation. b We want to show E is ecreasing in time. This is eqivalent to show E. t We have that E F t t t L,, t,, t where. Compting the total erivative as above we get L,, t,, t t L t t + L t + t L t + L t L t L 2. t L + t L 1 4 Where we se integration by parts an the fact that is constant in time at an 1, which meant the bonary term vanishe. Ths the energy E ecreases in time.