Variational Problems

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1 Vritionl Problems Com S 477/577 Notes Yn-Bin Ji Dec 7, 017 ThevritionlderivtiveoffunctionlJ[y]cnbedefinedsδJ/δy = F y x,y,y d dx F y x,y,y [1, pp. 7 9]. Euler s eqution essentilly sttes tht the vritionl derivtive of the functionl must vnish t n extremum. This is nlogous to the well-known result from clculus tht the derivtive of function must vnish t n extremum. 1 Vrible End Point Problem In this section, we consider simple cse of the vrible end point problem, which is stted s follows: Among ll curves whose end points lie on two verticl lines x = nd x = b, find the curve for which the functionl J[y] = Fx,y,y dx 1 hs n extremum. We determine the vrition of the functionl 1, which is the liner component of the increment J = J[y +h] J[y] = Fx,y +h,y +h Fx,y,y dx, due to n increment of h in y. The Tylor expnsion immeditely leds to δj = F y h+f y h dx. Unlike the fixed end point problem, the function hx no longer vnishes t the points x = nd x = b. Integrtion by prts now yields δj = F y d dx F y dx+f y hx x=b x= = F y d dx F y dx+f y x=b hb F y x= h. Consider ll functions hx with h = hb = 0 first. The condition δj = 0 implies tht F y d dx F y = 0. The mteril is dpted from the book Clculus of Vritions by I. M. Gelfnd nd S. V. Fomin, Prentice Hll Inc., 1963; Dover,

2 This mens tht the solution y of the end point problem must be solution of Euler s eqution. Suppose y is solution. The integrl in must therefore vnish, reducing δj = 0 to F y x=b hb F y x= h = 0. Since h is rbitrry, it follows from the bove eqution tht F y x= = 0 nd F y x=b = 0. 3 In summry, to solve the vrible end point problem, we first find generl solution to Euler s eqution, nd then use the conditions 3 to determine the constnts in the generl solution. Exmple 1. Strting t the origin, prticle slides down curve in the verticl plne. Find the curve such tht the prticle reches the verticl line x = b, where b 0, in the shortest time. The velocity v of motion equls gy from the conservtion of the totl energy, where g < 0 is the grvittionl ccelertion. Menwhile, it is lso determined long the curve s from which we immeditely hve v = ds dt = dx 1+y dt, 1+y dt = dx = v The bove gives us the trnsit time s functionl J = 0 1+y gy dx. 1+y gy dx. The integrnd F = 1+y / gy does not depend on x. So this is Cse where Euler s eqution cn be reduced to F y F y = C. 4 Since severl steps from Euler s eqution 4 led to F y = 1 y, gy 1+y y = y, for some < 0. y Given the downwrd sliding motion, we hve tht y 0, thus y y =, y which leds to Next, mke the substitution y = sin θ y dx = y dy. over [0,π] with dy dθ = sin θ cos θ 1 = sin θ cos θ.

3 Under the bove nd the substitution, sin θ dx = sin θ cos θ cos θ dθ = sin θ dθ 1 cosθ = dθ = r1 cosθdθ, where constnt r = / is introduced in the lst step. The solution to Euler s eqution 4 is fmily of cycloids x = rθ sinθ+c, y = r1 cosθ. Since the curve psses through the origin, the constnt c = 0. The first boundry condition F y x=0 = 0 in 3, which is essentilly y 0 = 0, is thus stisfied. To determine the vlue of r, we pply the second boundry condition F y x=b = 0, which reduces to y = 0 t x = b, i.e., 0 = dy dx = dy/dθ dx/dθ = rsinθ r rcosθ = sinθ 1 cosθ. The bove condition sttes tht the tngent to the curve t its right end point must be horizontl. Hence θ = π. Substituting this vlue into rθ sinθ = b, we obtin r = b π. Hence the curve trjectory is x,y = b θ sinθ,cosθ 1. π Figure 1 shows the optiml trjectory of the sliding prticle to rech the line x = π, represented by the cycloid θ sinθ,cosθ 1 over [0,π]. π Figure 1: The cycloid θ sinθ,cosθ 1 with 0 θ π is the minimum time trjectory long which prticle strting t the origin slides pss the verticl line x = π under grvity. The Cse of Severl Vribles Mny problems involve functionls tht depend on functions of severl independent vribles, for exmple, surfces in 3D dependingon two prmeters. Here we only look t how the solution to the 3

4 cse of single-vrible vritionl problems would crry over to the cse of functionls depending on surfces. We focus on the cse of two independent vribles but refer to [1] for the cse of more thn two vribles. Let Fx,y,z,p,q be twice continuously differentible with respect to ll five vribles, nd consider J[z] = Fx,y,z,z x,z y dxdy, 5 where is some closed region, nd z x nd z y re the prtil derivtives of zx,y with respect to x nd y. We re looking for function zx,y tht 1. twice continuously differentible with respect to x nd y in ;. ssumes given vlues on the boundry Γ of ; 3. yields n extremum of the functionl 5. Theorem 1 in the notes Clculus of Vritions does not depend on the form of the functionl J. Therefore, necessry condition for the functionl 5 to hve n extremum is tht its vrition vnishes. We must first clculte the vrition δj. Let hx, y be n rbitrry function with continuous first nd second derivtives in the region nd vnishes on it boundry Γ. Then, if the surfce zx,y stisfies conditions 1 3, so does zx,y+hx,y. We pply the Tylor series to the chnge in the functionl s follows: J = J[z +h] J[z] = Fx,y,z +h,z x +h x,z y +h y Fx,y,z,z x,z y dxdy = F z h+f zx h x +F zy h y dxdy +. 6 Menwhile, we hve tht F zx h x +F zy h y dxdy = x F z x h+ y F z y h = F zx hdy F zy hdx Γ dxdy x F z x + y F z y where the lst step used Green s theorem: Q x P dxdy = P dx+qdy. y Γ x F z x + y F z y hdxdy hdxdy. 7 In 7, the integrl long Γ is zero since h vnishes on the boundry. Substituting 7 into 6, we obtin δj = F z x F z x y F z y hx,ydxdy. 8 4

5 Thus, the condition δj = 0 implies tht the double integrl bove vnishes for ny hx,y with continuous derivtives up to the second order nd vnishing on the boundry Γ. This leds to the second-order prtil differentil eqution below 1, lso known s Euler s eqution: F z x F z x y F z y = 0. Next, we generlize the functionl 5 to llow second order prtil derivtives of z to pper in the integrnd function F. More specificlly, let Fx,y,z,p,q,r,s,t be function with continuous first nd second prtil derivtives with respective to ll rguments. The function z = zx, y hs continuous derivtives up to the fourth order, nd hs given vlues on the boundry Γ of. The functionl J[z] = Fx,y,z,z x,z y,z xx,z xy,z yy dxdy hs n extremum t z = zx,y only if the following version of Euler s eqution holds: F z x F z x y F z y + x F z xx + x y F z xy + y F z yy = 0. 9 Exmple. Surfce z = zx,y of lest re spnned by given contour. The problem reduces to finding the minimum of the functionl J[z] = 1+zx +z y dxdy, for which Euler s eqution hs the form z xx 1+zy z xyz x z y +z yy 1+zx = To understnd the geometric mening of eqution 10, we note tht the men curvture of the surfce is given s EN FM +GL H = EG F, where E,F,G rethe coefficients ofthe first fundmentl form ofthe surfce, nd L,M,N rethe coefficients of its second fundmentl form. We obtin tht nd hence, E = 1+z x, F = z x z y, G = 1+z y, L = z xx, M = 1+z x +zy z xy, N = 1+z x +zy z yy, 1+z x +zy H = z xx1+z y z xyz x z y +z yy 1+z x 1+z x +z y 3/. Thus, Euler s eqution 10 implies tht the men curvture of the solution surfce is zero everywhere. A surfce with zero men curvture everywhere is thus clled miniml surfce. Figure plots miniml surfce ctenoid cosucoshv/,sinucoshv/,v for 0 u π, 4 v 4. 1 To be rigorous, simple proof cn be derived from 8 nd tht h is rbitrry under stipulted conditions. 5

6 Figure : Ctenoid cosucoshv/,sinucoshv/,v over [0,π] [ 4,4]. 3 Severl Unknown Functions Let Fx,y 1,...,y n,z 1,...,z n be function with continuous first nd second derivtives with respect to ll its rguments. We now derive necessry conditions for n extremum of the functionl J[y 1,...,y n ] = Fx,y 1,...,y n,y 1,...,y ndx, where y 1 x,...,y n x re continuously differentible nd stisfy the boundry conditions y i = i, y i b = b i, for i = 1,...,n. Nmely, we re looking for n extremum of J defined on the set of smooth curves joining two points, 1,..., n nd b,b 1,...,b n in the n+1-dimensionl spce n+1. Similr to the derivtion of Euler s equtions for the vritionl problems studied before, we dd continuously differentible function h i x to ech y i x such tht the resulting y i x+h i x still stisfy the boundry conditions. Therefore, h i = h i b = 0, for i = 1,...,n. The vrition of J[y 1,...,y n ] is found to be δj = n F yi h i +F y i h idx. i=1 At ech step, settingll butoneoftheh i tozero, weobtinthefollowingsystemof Euler sequtions tht must be stisfied t n extremum: F yi d dx F y i = 0, i = 1,...,n. 11 6

7 Exmple 3. Geodesics. Given prmetric surfce σ = σu,v, the vritionl problem here is to find the minimum distnce between p nd q on the surfce. Prmetrize curve from p nd q s αt = σut,vt so tht p = α nd q = αt 1. We denote by. differentition with respect to t. The rc length between the two points is J[u,v] = E u +F u v +G v dt, 1 where E,F,G re the coefficients of the first fundmentl form of the surfce given s E = σ u σ u, F = σ u σ v, G = σ v σ v. Euler s equtions for the functionl 1 become E u u +F u u v +G u v E u +F u v +G v d E u+f v dt E u +F u v +G v E v u +F v u v +G v v E u +F u v +G v d F u+g v dt E u +F u v +G v = 0, 13 = Since the prmetriztion does not chnge the totl length of the curve, we my use the rc length prmetriztion. From surfce geometry, we know tht this mens Then equtions 13 nd 14 reduce to E u +F u v +G v = 1. d dt E u+f v = 1 E u u +F u u v +G v v, d dt F u+g v = 1 E v u +F v u v +G v v. They re essentilly the geodesic equtions. This proves wht we hve lerned before tht the shortest curve on σ connecting two surfce points p nd q is geodesic between the two points. Sometimes, it is more convenient to use one of the surfce prmeters to prmetrize the geodesic. In such cse, we still need to fll bck on equtions 13 nd 14. To illustrte, we now find the geodesics of the circulr cylinder σ = cosφ,sinφ,z. First, we obtin the coefficients of the first fundmentl form: E =, F = 0, G = 1. Substitute these coefficients into 13 nd 14: d φ dt φ = 0, +ż d ż dt φ = 0, +ż 7

8 Integrte the two equtions bove: φ φ +ż = C 1, ż φ +ż = C. Dividing the lst eqution by the one before, we obtin which hs the solution dz dφ = D 1, z = D 1 φ+d. The geodesic between two points on cylinder thus is helix lying on the cylinder. Given two points σφ 0,z 0 nd σφ 1,z 1, the helix is described s αφ = cosφ,sinφ, z 0 z 1 φ+ φ 0z 1 φ 1 z 0. φ 0 φ 1 φ 0 φ 1 Figure 3 shows geodesic cosφ,sinφ,φ connecting two points with φ 0 = 0 nd φ 1 = 3 cosφ,sinφ,z which is plotted over the subdomin [0,π] [ 1,4]. on cylinder Figure 3: A geodesic on cylinder. 4 Vritionl Problems in Prmetric Form We hve considered functionls of curves given by equtions of the form y = yx. Often, it is more convenient to consider functionls of curves in prmetric form, sy, xt, yt. Given functionl 8

9 of such curve x1 Fx,y,y dx, 15 x 0 over [,t 1 ] such tht x = x 0 nd xt 1 = x 1, we cn rewrite it s xt,yt, ẏt ẋdt = Φx,y,ẋ,ẏdt. ẋt F Here the overdot denotes differentition with respect to t. This becomes functionl depending on two unknown functions xt nd yt. The function Φ does not hve t s n explicit rgument. It is positive-homogeneous of degree 1 in ẋ nd ẏ: for ny c > 0. Conversely, let Φx,y,cẋ,cẏ cφx,y,ẋ,ẏ, Φx,y,ẋ,ẏdt befunctionlwhoseintegrndφispositive-homogeneous ofdegree1inẋndẏ. Wenowshowtht the vlue of the functionl is independent of the curve prmeteriztion. Suppose we reprmetrize the curve with τ over [τ 0,τ 1 ] such tht t = tτ, nd dt/dτ > 0 over [τ 0,τ 1 ]. Then τ1 τ 0 Φ x,y, dx dτ, dy dτ dτ = = = τ1 Φ τ 0 x,y,ẋ dt dτ,ẏdt dτ Φx,y,ẋ,ẏ dt dτ dτ Φx,y,ẋ,ẏdt. dτ Now, suppose the curve y = yx hs prmeteriztion in t tht reduces the functionl 15 to the form Φx,y,ẋ,ẏdt. Applying Euler s eqution 11, we end up with two equtions: Φ x d dt Φ ẋ = 0, Φ y d dt Φ ẏ = We cn solve them for x nd y. The two equtions in 16 must be equivlent to the single Euler eqution: F y d dx F y = 0, which results from the vritionl problem for the originl functionl 15. Hence the two equtions re not independent. We refer to[1] for other vritionl problems such s involving higher derivtives, with subsidiry conditions, involving multiple integrls, etc. Also, we refer to [] for ppliction of the clculus of vritions in theory of elsticity, quntum mechnics, nd electrosttics. 9

10 eferences [1] I. M. Gelfnd nd S. V. Fomin. Clculus of Vritions Prentice-Hll, Inc., 1963; Dover Publictions, Inc., 000. []. Weinstock. Clculus of Vritions: With Applictions to Physics nd Engineering. Dover Publictions, Inc.,