The Basic Functional 2 1


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1 2 The Bsic Functionl 2 1
2 Chpter 2: THE BASIC FUNCTIONAL TABLE OF CONTENTS Pge 2.1 Introduction The First Vrition The Euler Eqution Expnded Form of the EulerLgrnge Eqution Dul Form Degenerte Cses Legendre s Test Nturl BoundryConditions EulerLegendre Eqution Vnishes Identiclly Jcobi s AccessoryEqution VF, WF nd SF For Bsic Functionl Exmples Shortest Distnce Between Two Points Brchistochrone: Curve Of Shortest Descent
3 2.3 THE EULER EQUATION 2.1. Introduction This Chpter tkes up the bsic onedimensionl functionl (1.3) introduced in to illustrte key opertion of Vritionl Clculus: getting vritions. A necessry condition for the functionl to be sttionry (in prticulr, to ttin minimum or mximum) is tht its first vrition vnish. Severl exmples of this procedure re given to illustrte pplictions. The functionl (1.3) nd end conditions reproduced below for convenience of the reder: J[y] = b F ( y(x), y (x), x ), x = [, b], b, y() =ŷ, y(b) =ŷ b. (2.1) Here y = y(x) is C 1 singlevlued rel function of x over the intervl [, b]. The vritionl form δ J = 0 is worked out to illustrte the derivtion of the EulerLgrnge equtions, s well s the concept of strong nd wek forms The First Vrition A necessry condition for functionl (2.1) to be sttionry with respect to its input function is tht its first vrition vnishes δ J = 0. (2.2) Let y(x) nd y(x) + δy(x) be two dmissible input functions. By stipultion the difference δy(x) is lso dmissible. This will be clled the vrition of y(x). As defined δy(x) is not necessrily infinitesiml but finite increment. 1 To tke it to the infinitesiml limit, scle it s δy(x) = ɛ h(x), δy(x) = ɛ h (x), (2.3) in which ɛ [0, 1] is scling prmeter nd h is reference function independent 2 of ɛ. Insert into the functionl to express its first vrition s 1 b [ ( δ J = lim F y(x) + ɛh(x), y (x) + ɛh (x), x ) F ( y(x), y (x), x )] dx. (2.4) ɛ 0 ɛ 1 The literture is divided s regrds nottion nd priori ssumptions on δy(x). For exmple [321], uses different symbol, sy h(x), for the finite vrition nd ɛ h(x) for the infinitesiml. Others renme the infinitesiml limit s δy(x). Others simply restrict δy(x) to be infinitesiml from the strt. The nottion δ J, δ 2 J, etc., for the functionl vritions is universl. 2 In mthemticl rigorous tretments of Vritionl Clculus, (2.3) is clled wek vrition. Its key feture is tht both δy(x) nd δy (x) go to zero s ɛ 0. In the more restrictive strong vritions, δy (x) does not necessrily go to zero. For the problems discussed in this book, the distinction is irrelevnt. 2 3
4 Chpter 2: THE BASIC FUNCTIONAL 2.3. The Euler Eqution Expnding δ J in powers of ɛ s Mclurin series: 1 b [ δ J = lim ɛ ɛ 0 ɛ y h(x) + ɛ ] b y h (x) dx + O(ɛ 2 ) = δf dx+ O(ɛ 2 ). (2.5) in which δf = [ y h(x) + ] y h (x). (2.6) is clled the first vrition of the integrnd F with respect to h(x). Notice tht the scling prmeter ɛ hs disppered from it. Integrting by prts the second term in the integrl (2.5) yields b b y h (x) dx = d dx y h(x) dx + y h(x) b 0. (2.7) The crossed out term vnishes becuse h() = h(b) = 0 on ccount of the essentil BC there. Replcing into (2.5) llows the extrction of h(x) s fctor: δ J = b [ y d ] b h(x) dx = E h(x) dx = 0. (2.8) dx y in which E = y d = F dx y y (F y ). (2.9) Since h(x) is rbitrry, the fundmentl lemm of Vritionl Clculus (see, e.g., Chpter 1 of [321]) tells us tht for δ J = 0, the differentil form (2.9) must vnish identiclly: E = 0 x [, b]. (2.10) The differentil eqution (2.10) is clled the Euler eqution or EulerLgrnge eqution ssocited with the bsic vritionl functionl. (Some older books cll this chrcteristic eqution, nme now reserved for other expressions.) Remrk 2.1. The ODE (2.10) supplies only necessry condition for n extremum of J[y]. A solution my provide mximum, minimum, or simply sttionry vlue tht is neither. The derivtion nd checking of sufficient conditions for mximum or minimum is much more involved. Conditions y() = y t x = or y(b) = y b t x = b re clled essentil boundry conditions. 2 4
5 2.3 THE EULER EQUATION Expnded Form of the EulerLgrnge Eqution Since / y is generlly lso function of x, y nd y, differentiting the second term of E with respect to x gives E = y 2 F y 2 F y 2 F = F x y y y y y y F xy y F yy y F y y. (2.11) Consequently E = 0 is generlly secondorder ordinry differentil eqution (ODE). Its generl solution hs two rbitrry coefficients. Those coefficients re determined from the end conditions y() =ŷ nd y(b) =ŷ b, tking one from ech end. A solution of E = 0 tht stisfies the end BC is clled n extremizing function or simply extreml. As noted in Remrk 2.1, the functionl J does not necessrily ssumes mximum or minimum vlue t n extreml, lthough tht cn be often checked by direct rguments Dul Form Differentiting F = F(y(x), y (x), x) with respect to x nd using the chin rule we get df dx = x + y y + y y = F x + F y y + F y y. (2.12) The second term of E cn be expnded s ( d y ) = d ( ) + y. (2.13) dx y dx y y Substrcting (2.13) from (2.12) nd collecting terms we get ( d F ) y ( dx y x = y d ) y = E y. (2.14) dx y But since E vnishes, so does the RHS of (2.14). This gives nother form of the EulerLgrnge eqution, which we will denote by Ē: Ē = d ( F ) y dx y x = ( F F y y ) Fx. (2.15) This receives the nme Legendre trnsform or dul form of the EulerLgrnge eqution in the literture. It plys n importnt role in some pplictions, such s Hmiltonin dynmics Degenerte Cses If F is independent of x, then / x = 0 nd the dul form (2.15) reduces to (F F y y ) = 0, which hs the first integrl F y y = C. (2.16) 2 5
6 Chpter 2: THE BASIC FUNCTIONAL in which C is constnt of integrtion. In this cse the EulerLgrnge eqution reduces to firstorder ODE. Another integrtion then provides the second constnt. If F is independent of y, then / y = 0. The EulerLgrnge eqution (2.10) reduces to d(/ y )/dx = 0, which immeditely integrtes to y = C. (2.17) This is firstorder ODE tht involves x nd y only. If F is independent of x nd y, the EulerLgrnge eqution (2.10) integrtes immeditely to y = C 1. Another integrtion gives the liner solution y = C 1 x + C 2, (2.18) in which constnts C 1 nd C 2 cn be determined from the end conditions. If F is independent of y, then / y = 0 nd the EulerLgrnge eqution (2.10) reduces to / y = 0. Consequently F = F(x). (2.19) Tht is, the integrnd must be function of x only. The resulting lgebric eqution is generlly incomptible with the end conditions Legendre s Test If the rnge of integrtion [, b] is sufficiently smll nd if the sign of 2 F/ y 2 is constnt throughout this rnge then when δ J = 0 the vlue of J is mximum (minimum) if tht sign is negtive (positive). For rbitrry [, b] Legendre s test is necessry condition for n extremum but is not sufficient. This point is discussed further in Chpter 5 of GF Nturl Boundry Conditions Conditions F y = 0tx = or F y = 0tx = b re clled nturl boundry conditions. Four combintions re possible. See exmple in Lnczos, pge 68 (where br ctully mens plne bem ) EulerLegendre Eqution Vnishes Identiclly If E vnishes identiclly in [, b], then J cn be evluted s function of boundry vlues Jcobi s Accessory Eqution Let y = ŷ(x) be the extreml tht psses though given end vlues y() = y nd y(b) = y b. Substitute this vlue into F(y, y, x) nd let F 00 = 2 F/ ŷ 2, F 01 = 2 F/ ŷ ŷ nd F 11 = 2 F/ ŷ 2. Then Jcobi s ccessory eqution is ( F 00 d ) dx F 01 u d ( ) du F 11 = 0 dx dx (2.20) This topic is covered in detil in Chpter 5 of GelfndFomin. 2 6
7 2.8 VF, WF AND SF FOR BASIC FUNCTIONAL Figure 2.1 Sketch of vritionl, wek nd strong forms for the bsic functionl (6.1) 2 7
8 Chpter 2: THE BASIC FUNCTIONAL 2.8. VF, WF nd SF For Bsic Functionl Figure 2.1 sketches the reltion between Vritionl Form, Wek Form nd Strong Form for the bsic functionl (6.2). The importnt role of h(x) in the construction of the Wek Form (WF) is pprent. When WF is constructed from the SF, h(x) is often clled the weight function or test function nd is denoted by w or ψ or ϕ. Sometimes it is convenient to regrd h s Lgrnge multiplier field in which cse it is usully denoted by λ. Only if the WF cn be connected to VF is h interpretble s the vrition of function y(x) Exmples All of the following exmples pertin to the bsic functionl Shortest Distnce Between Two Points Problem. Two given points, lbeled A(x A, y A ) nd B(x B, y B ) re locted on plne {x, y}. Find the shortest distnce between A nd B. Solution. Admissible functions y(x) pss through A nd B, nd re ssumed to be singlevlued nd of clss C 1. The distnce to be minimized is d AB = B A ds = xb x A 1 + (y ) 2 dx. (2.21) The integrnd F = 1 + (y ) 2 does not depends on y. Thus its EL eqution reduces to the firstorder ODE (2.17), which becomes y = k, constnt. (2.22) 1 + (y ) 2 Therefore y = k/ k 2 1, which shows tht y is constnt, sy C 1. Integrting y = C 1 we get y(x) = C 1 x + C 2. (2.23) This is the eqution of stright line. The constnts of integrtion re determined by mking the line pss through A nd B. If the points re distinct, the line is unique. To check whether the solution (2.23) extremizes d AB, observe tht F y y = 2 F y y = [ ] (y ) 2 1 y y = (1 + C 2 = R, constnt. (2.24) 1 )3/2 y C 1 If the + sign of the squre root is tken, R > 0 over the problem domin x [x A, x B ]. Thus the Legendre test sys tht (?) mkes d AB minimum. This is cler from the physicl interprettion. 2 8
9 2.9EXAMPLES Brchistochrone: Curve Of Shortest Descent The following problem, first solved by John Bernoulli in 1696, led to the formultion of Vritionl Clculus by Euler nd Lgrnge in its current form. Problem. Two points locted on the plne {x, y} re given: A(x A, y A ) nd B(x B, y B ), with y B < y A. They re joined by smooth curved pth c s pictured in Figure?. Constnt grvity g cts in the y direction. A prticle (point mss) is relesed from A t time t = 0, with zero initil velocity. It slides without friction long the pth, nd rrives t B t time t = T BA. Find the plne curve y = y(x) tht minimizes T BA. Solution. The function y = y(x) tht defines the pth is ssumed to be of clss C 1 nd singlevlued. Let P(x, y, t) denote the position of the prticle t time t, It speed is v = ds/dt = v(x, y, t) The prticle energy is the sum of the kinetic energy T = 1 2 m v2 nd its potentil energy V = mg(y y A ). Since T + V is conserved under zero friction,solving for v gives v = 2g(y y A ), which is independent of the mss. For simplicity tke y A = 0; then v = sqrt2g ybut v = ds/dt with ds = 1 + (y ) 2 dx. The time differentil is dt = ds/v, which integrted from A tp B yields the descent time: B x=xb 1 + (y T BA = dt = ) 2 dx. (2.25) A x=0 2gy Since the integrnd is independent of x one cn immeditely obtin first integrl of the EL eqution, which cn be simplified to y(1 + (y ) 2 ) = 2R, R constnt. (2.26) To solve this ODE, it is convenient to pss to prmetric form by putting y = cot φ/2. It cn be then verified tht the generl solution tht psses through x A = y A = 0is x = R(φ sin φ), y = R(1 cos 2φ). (2.27) This represents oneprmeter fmily of cycloids with R s the rdius of the rolling circle nd 2φ s the ngulr prmeter. The vlue of R is determined by mking the pth pss through point B. The generting circle rolls on the horizontl line through A, tht is, y = 0. See Figure. The Legendre test indictes tht the solution pth minmizes T BA for sufficiently smll distnces between A nd B. 2 9
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