Calculus of Variations

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1 Clculus of Vritions Com S 477/577 Notes) Yn-Bin Ji Dec 4, Introduction A functionl ssigns rel number to ech function or curve) in some clss. One might sy tht functionl is function of nother function or curve). Functionls hve n importnt role in mny problems rising in nlysis, optimiztion, mechnics, geometry, etc. Exmple 1. Consider ll possible pths joining two points p nd q in the plne, nd prticle moving long ny of these pths with velocity vx,y) t the point x.y). Then we obtin functionl tht ssigns to ech pth the time of trversl by the prticle. Exmple 2. Let yx) be n rbitrry continuously differentible function defined on the intervl [, b]. Then the integrl is functionl on the set of ll such functions yx). y 2 x)dx Exmple 3. Let Fx,y,z) be continuous function of three vribles. Then the integrl Fx,yx),y x))dx, where yx) rnges over the set of ll continuously differentible functions defined over the intervl [, b], is functionl. Different choices of the function Fx, y, z) give different functionls. For exmple, if then J[y] is the length of the curve yx). If Fx,y,z) 1+z 2, Fx,y,z) z 2, J[y] reduces to the cse considered in the previous exmple. Instnces of problems involving functionls were considered more thn three hundred yers go, nd the first importnt results re due to Euler ). Though still not hving methods The mteril is dpted from the book Clculus of Vritions by I. M. Gelfnd nd S. V. Fomin, Prentice Hll Inc., 1963; Dover,

2 comprble to those of clssicl nlysis in generlity, the clculus of functionls hs hd the most development in finding the extrem of functionls, brnch referred to s the clculus of vritions. Listed below re few typicl vritionl problems. 1. Find the shortest plne curve connecting two points p nd q, tht is, find the curve y yx) for which the functionl 1+y 2 dx chieves its minimum. The shortest curve turns out to be the line segment pq. 2. Let p nd q be two points in the verticl plne 1. Consider prticle sliding under grvity long some pth curve) joining p nd q. The time it tkes the prticle to rech q from p depends on the curve, nd hence is functionl. The curve tht results in the lest time is clled the brchistochrone. The brchistochrone problem ws posed by John Bernoulli in 1696, nd solved by himself, Jmes Bernoulli, Newton, nd L Hospitl. The brchistochrone turns out to be cycloid, s we will discuss lter. 3. The isoperimetric problem ws solved by Euler: Among ll closed plne curves of given length, find the one tht encloses the gretest re. The nswer turns out, not surprisingly, to be circle. All of the bove problems cn be written in the form Fx,y,y )dx. Such function hs locliztion property tht the vlue of the functionl equls the sum of its vlues over segments generted by dividing the curve y yx). Below is functionl exmple tht does not hve this property: x 1+y 2 dx 1+y 2 dx. It is importnt to see how problems in clculus of vritions re relted to those of clssicl nlysis, especilly, to the study of functions of n vribles. Consider functionl of the form Fx,y,y )dx, y) y, yb) y b. To see the connection, we divide the intervl [,b] into n+1 equl prts: x 0,x 1,...,x n,x n+1 b. Then, replce the curve y yx) by the polygonl line with vertices x 0,y ),x 1,yx 1 ),...,x n,yx n )),x n+1,y b ). Essentilly, we re pproximting the functionl J[y] by the sum n+1 Jy 1,...,y n ) F i1 1 The two points re ssumed to hve distinct x-coordinte vlues. x i,y i, y ) i y i 1 h, h 2

3 where y i yx i ) nd h b n+1. Thus, we cn regrd the vritionl problem s tht of finding the extrem of the function Jy 1,...,y n ), with n. In this sense, functionls cn be regrded s functions of infinitely mny vribles, nd the clculus of vritions s the corresponding nlog of differentil clculus. 2 The Vrition of Functionl In vritionl problem, we tret ech function belonging to some clss s point in some spce referred to s the function spce. For instnce, if we re deling with functionl of the form Fx,y,y )dx, the spce includes ll functions with continuous first derivtive. In the cse of functionl Fx,y,y,y )dx, the pproprite function spce is the set of ll functions whose second derivtives re continuous. For vritionl problem, we consider the liner spce of fesible functions. We need to mesure the closeness between two elements in function spce. This is done by introducing the concept of the norm of function. 2 Let the spce D, or more precisely D,b), consist of ll continuous functions defined on n intervl [, b]. Addition nd multipliction by rel num- y bers re in the usul sense. The norm is defined yx) s y 0 mx yx). 2ǫ x b As illustrted in Figure 1, the distnce between the functions y x) nd yx) does not exceed ǫ if the grph of the function yx) lies inside bnd of width 2ǫ centered t the grph of y x). Similrly, the spce D n, or more precisely D n,b), consists of ll functions defined on [,b] tht re n times continuously differentible. The norm now is defined s n y n mx y i) x). x b i0 y x) Figure 1: Distnce between two functions is the mximum bsolute difference of their vlues t ny point. A functionl J[y] is sid to be continuous t point i.e., function) y 0 in normed function spce F if for ny ǫ > 0, there exists δ > 0 such tht J[y] J[y 0 ] < ǫ, for ll y y 0 < δ. A continuous functionl J is sid to be liner if 2 We hve indeed used the concept of the norm in defining the best pproximtion of function using fmily of functions. b x 3

4 1. J[αh] αj[h] for ny function h F nd α R; 2. J[h 1 +h 2 ] J[h 1 ]+J[h 2 ] for ny h 1,h 2 F. Exmple 4. The integrl defines liner functionl on D, b). J[h] hx) dx Exmple 5. The integrl J[h] ) α 0 x)hx)+α 1 x)h x)+ +α n x)h n) x) dx, where α i x) re functions in D,b), defines liner functionl on D n,b). Let J[y] be functionl defined on some normed liner spce, nd let J[h] J[y +h] J[y] be its chnge due to n increment hx) of the vrible y yx). When yx) is fixed, J[h] is functionl of h. Suppose J[h] φh)+ǫ h, where φh) is liner functionl nd h 0 s ǫ 0. Then the functionl J[y] is sid to be differentible. The liner functionl φ[h] which differs from J[h] by the higher order infinitesiml term ǫ h, is clled the vrition or differentil) of J[y] nd denoted by δj[h]. This differentil is unique when the functionl is differentible. Theorem 1 The differentible functionl J[y] hs n extremum t y y only if its vrition vnishes for y y, tht is, δj[h] 0 for ll dmissible functions h. In the theorem, dmissible functions refer to those tht stisfy the constrints of given vritionl problem. We refer to [1, p. 13] for proof of Theorem 1. The function J[y] hs n extremum for y y if J[y] J[y ] does not chnge its sign in some neighborhood of the curve y x). 3 Euler s Eqution Now we consider wht might be clled the simplest vritionl problem. Let Fx,y,z) be function which is twice continuously differentible with respect to ll rguments. Among ll functions yx) tht re continuously differentible over [, b] nd stisfy the boundry conditions y) y nd yb) y b, 1) 4

5 find the one for which the functionl Fx,y,y )dx 2) hs n extremum. Suppose we increse yx) by hx). For yx) + hx) to continue to stisfy the boundry conditions 1), we must hve h) hb) 0. 3) The corresponding increment in the functionl is J J[y +h] J[y] Fx,y +h,y +h )dx Fx,y,y )dx ) Fx,y +h,y +h ) Fx,y,y ) dx F y x,y,y )h+f y x,y,y )h ) dx+, fter pplying Tylor s expnsion. The integrl in the right hnd side of the lst eqution bove represents the principl liner prt of the increment J, hence the vrition of J[y] is δj F y x,y,y )h+f y x,y,y )h ) dx. According to Theorem 1, J[y] hs n extremum only if δj 0. Thus we hve 0 F y x,y,y )h+f y x,y,y )h )dx h F y d ) dx F y dx+f y h b h F y d ) dx F y dx, under the conditions 3). Since the function h is rbitrry except h) hb) 0, it is not difficult to show tht the lst eqution bove implies F y d dx F y 0. 4) Eqution 4) is clled Euler s eqution. Thus, the functionl 2) hs n extremum for given function yx) only if yx) stisfies Euler s eqution. Since the eqution is second order, its solution depends on two constnts, which cn be determined from the two boundry conditions 1). Nevertheless, in some specil cses, Euler s eqution cn be reduced to first-order differentil eqution. 5

6 Cse 1 The integrnd does not depend on x. So the functionl hs the form Euler s eqution becomes Tke the prtil derivtive: Fy,y )dx. F y d dx F y y,y ) 0. F y F y yy F y y y 0. Multiplying the bove eqution by y, we obtin F y y F y yy 2 F y y y y d dx F y F y ) 0. Thus, we end up with first order differentil eqution: F y F y C, for some constnt C. Cse 2 The integrnd does not depend on y. So the functionl hs the form Euler s eqution reduces to which yields the first-order eqution: Fx,y )dx. d dx F y 0, F y C, for some constnt C. Solving this eqution for y, we obtin n eqution of the form y fx,c), which is integrted for the curve: y fx,c)dx+d, for some constnt D. Cse 3 The integrnd does not depend on y. Euler s eqution tkes the form which is not differentil eqution. F y x,y) 0, 6

7 Cse 4 In vriety of problems, the functionl is the integrl of function fx,y) with respect to the rc length 1+y 2 dx. Here F fx,y) 1+y 2. Euler s eqution cn be trnsformed s follows: 0 F y d dx F y f y x,y) 1+y 2 d dx y fx,y) 1+y 2 f y 1+y 2 y f x f y 2 y f y 1+y 2 1+y 2 1+y 2 ) 3/2 1 f y f x y y ) f 1+y 2 1+y 2, ) tht is, y f y f x y f 0. 1+y 2 Exmple 6. For the functionl Euler s eqution reduces to the eqution x y) 2 dx, x y 0. So the solution is the line y x, long which the integrl vnishes. This is Cse 3. Exmple 7. Consider the functionl y 2 dx, y1) 0, y2) 1. x The integrnd does not contin y. This is Cse 2, nd Euler s eqution hs the form F y C, for some C. Thus, so tht y x C. 1+y 2 y 2 1 C 2 x 2 ) C 2 x 2. Since C cn tke on either positive or negtive vlue, we obtin the derivtive from the bove: y Cx 1 C2 x 2. Integrting the bove gives us y Cx 1 C2 x 2 dx 1 C 1 C2 x 2 +D, or, equivlently, y D) 2 +x 2 1 C 2. 7

8 Nmely, the solution is circle centered on the y-xis. From the boundry conditions, we find tht C 1 5 nd D 2. Thus, the finl solution is y 2) 2 +x 2 5. Exmple 8. Among ll the curves joining two points x 0,y 0 ) nd x 1,y 1 ), find the one which genertes the surfce of minimum re when rotted bout the x-xis. In this problem, the functionl is the re of the surfce of revolution s result of rotting the curve y yx) bout the x-xis: 2π x1 x 0 y 1+y 2 dx. We let Fy,y ) y 1+y 2 nd minimize the integrl. This is cse 1 where Euler s eqution yields the first order eqution: F y F y C. Thus, we hve y 2 y 1+y 2 y C, 1+y 2 or So, which becomes with solution or equivlently, y C 1+y 2. y y2 C ± 2 C 2, dx C y2 C 2dy, x+d Cln y + y 2 C 2, C y Ccosh x+d C. The solution curve in Exmple 7 is ctenry, nd the surfce generted by rottion of the ctenry is clled ctenoid. The constnts C nd D re determined using the conditions yx 0 ) y 0 nd yx 1 ) y 1. Figure 2 plots the solution curve y 2cosh x+3 2 ) joining two points 1,2cosh2)) nd 2,cosh5 2 )). Depending on the loctions x 0,y 0 ) nd x 1,y 1 ), the solution surfce my not exist. Consider the cse where the distnce between the two points is sufficiently lrge compred to y 0 nd y 1. The re generted by rotting the polygonl line p 1 p 2 p 3 p 4, where p 1 x 0,y 0 ), p 2 x 0,0), p 3 x 1,0), nd p 4 x 1,y 1 ) will be less thn the re generted by rotting smooth curve pssing through the points. 8

9 Figure 2: A ctenry y 2cosh x+3 2 ) over [1,2]. References [1] I. M. Gelfnd nd S. V. Fomin. Clculus of Vritions Prentice-Hll, Inc., 1963; Dover Publictions, Inc., [2] R. Weinstock. Clculus of Vritions: With Applictions to Physics nd Engineering. Dover Publictions, Inc.,