6 Variational Methods

Transcription

1 6 Vritionl Methods 6. Introduction nd Motivtion from Clculus The min intention of this chpter is to provide the students with brief introduction to vrious methods in the clculus of vritions. We re minly interested in the mthemticl ide of minimiztion of function or more precisely functionl). The reson for the nme is tht in order to find minim we will use techniques which re refered to vritionl methods. In the study of functions of single vrible we consider the problem of finding locl minimum. Let f :[, b] R be rel vlued function defined on bounded intervl. By definition, f hs locl minimum t x if nd only if there exists neighborhood B ɛ x )={x : x x <ɛ} such tht fx) fx ) for ll x B ɛ x ) [, b], x x. If f hs locl minimum t x nd the derivtive f x) exists, then we hve two cses:. If x, b) i.e., x is n interior point), then 2. If x is boundry point on [, b], then f x )=. f x )x x ), for ll x [, b]. The second cse is equivlent to f x ) if x = f x ) if x = b). A centrl concept for extreml problems is tht of criticl point. If f x ) exists, then by definition f hs criticl point t x if nd only if f x )=, i.e., the tngent line is horizontl t x = x. The following re criticl points: locl mxim, locl minim, horizontl inflection points. Another importnt tool in the study of locl extrem of functions is the Tylor expnsion, provided tht f is differentible sufficient number of times. For exmple, if f x )=, f x ) >, then fx) =fx )+ f x )x x ) ! In this cse we hve strict locl minimum t x. For functions of severl vribles necessry condition for x R to be locl mximum or locl minimum of function f :Ω R n R is tht fx )= R n, where = x x 2. x n is the grdient.

2 Geometriclly, fx ) is the lrgest vlue tken on by fx) ner x, if for ny y nd α smll enough) we hve fx ) >fx + αy). Now using tylor s theorem we hve fx + αy) =fx )+α fx ) y + α 2 terms bounded by constnt. Now for smll α,if fx ), there must be direction y for which fx ) y. But then chnge of signs of α would chnge the sign of fx ) fx + αy) which is impossible. Therefore we must hve fx )=. The min problem in the clculus of vritions is to minimize or mximize) not functions but functionls. For exmple, suppose we hve function F x, y, z) nd we wish to find function yx) which minimizes Jy) = F x, yx),y x)) dx 6..) over the set of C functions, i.e., y C [, ] with y) = y nd y) = y. The clss of functions, C [, ] in this cse, is clled the clss of dmissible functions for this problem. Suppose now tht ỹ C [, ] minimizes J nd ηx) is ny dmissible function with η) = η)=.for ny sclr α,wehve J α ỹ + αη) α= =. By the chin rule this holds if nd only if = J α ỹ + αη) α= = y x, ỹ + αη), ỹ + αη )) α= η + ) dx = y x, ỹ + αη), ỹ + αη )) α= η [ y d )] ηdx+ dx y [ ] x= y η. 6..2) x= Here, to obtin the lst equlity, we hve ssumed tht it is justified, in the second term, to crry out n integrtion by prts. We hve then set α =. Assuming tht y) = y nd y) = y re prescribed, then we need η) = η)=. We obtin [ = y d )] η dx. dx y This must hold for ll dmissible η stisfying η) = η)=. At this point we obtin n interesting point in mthemticl history. Euler, using dubious rigor, concluded tht from the bove, we cn conclude y d =. 6..3) dx y Relizing tht Euler hd not crefully justified this step gve n lterntive derivtion by concluding tht if the bove integrl vnishes for ll dmissible η vnishing t the end points then the coefficient of η must vnish. He did not supply proof of this though nd ws in turn criticized by Euler for this shortcoming. It ws ll put to rest by Dubois Reymond who proved the following result clled the Fundmentl Lemm of the Clculus of Vritions. 2

3 Lemm 6.. If M C[, b] nd Mx)ηx) dx = for ll η S = {ϕ C [, b] :ϕ) = ϕ)=}, then Mx) = for ll x [, b]. by contrdiction. Suppose there exists n x, b) so tht Mx ). Then we cn, without loss of generlity, ssume tht Mx >. Since M C[, b], there exists B ɛ x ), b) such tht Mx) > for ll x B ɛ x ). Let { x x ɛ) 2 x x + ɛ) 2 for x B ɛ x ) ηx) = for x B ɛ x ). With this choice of dmissible η we hve Mx)ηx) dx = x x +ɛ) x x ɛ) Mx)x x ɛ) 2 x x + ɛ) 2 dx >, which contrdicts our hypothesis. Thus we hve Mx) =for ll x, b). By continuity we lso must hve M) =Mb) =. Remrk 6... The eqution 6..3) is clled the Euler-Lgrnge eqution. 2. A solution of the Euler-Lgrnge eqution is clled n extreml. 3. An extreml tht stisfies the end conditions is clled sttionry function. Such function still my or my not be mximum or minimum for j. 4. We note tht in the derivtives y nd y we tret x, y nd y s independent vribles. 5. Since is, in generl, function of x explicitly nd lso implicitly through y nd y, the y second term in 6..3) cn be written in expnded form s x ) + y y y Thus the Euler-Lgrnge eqution cn be eritten s ) dy dx + y y ) dy dx. d 2 y F y y dx + F dy 2 y y dx +F y x F y )=. 6..4) 6. The eqution is of second order in y unless F y y = 2 F y 2 =so tht, in generl, there re two constnts of integrtion vilble to stisfy the two end conditions. 3

4 7. The eqution 6..4) is equivlent to [ d F ) dy y dx y dx 8. There re some specil cses of 6..3) nd 6..5): ) From 6..5) we see tht if F does not depend explicitly on x, i.e., J = ] = 6..5) x F y, y ) dx then F y d dx F y = F y F y yy F y y y. Multiplying by y we hve F y y F y yy ) 2 F y y y y = d dx F y F y ). Thus we hve = F = c. x y b) From 6..5) we see tht if F does not depend explicitly on y, i.e., J = F x, y ) dx then In this cse Euler s eqution becomes d dx F y = which gives F y =where F does not contin y. This is first order Euler eqution nd solving for y we hve y = fx, c) which cn be solved by qudrture. Thus y = y = c. c) From 6..5) we see tht if F does not depend explicitly on y, i.e., J = F x, y) dx then Euler s eqution becomes F y x, y) = nd hence it is not differentil eqution but simply n eqution whose solution consists of one or more curves. 4

5 9. In mny problems the functionl J hs the specil form J = fx, y) +y ) 2 dx, representing the integrl of function fx, y) with respect to rclength s so ds = +y ) 2 dx). In this cse Euler s eqution cn be trnsformed into y d = f dx y x, y) [ ] +y ) 2 d y fx, y) dx +y ) 2 Thus we obtin y = f y +y ) 2 f x +y ) f y ) 2 2 y +y ) f y 2 + y ) 2 ) 3/2 = ) f y f x y y f. +y ) 2 + y ) 2 ) y f y f x y f + y ) 2 ) 6..6) 6.. Simple Exmples Let us begin with simple exmple Exmple 6.. Find the miniml surfce of revolution pssing through two given points x,y ), x 2,y 2 ) in the plne. This leds to minimizing the integrl J =2π x2 x y + y ) 2 ) /2 dx. 6..7) The surfce re S of smll surfce of revolution formed by revolving the curve y = yx) from x = x to x = x 2 bout the x-xis pproximtely equls the circumference 2yπ times the rclength of the curve from yx) to yx + x), denoted by s: S =2πy s. 5

6 Now recll tht infinitesiml rc length is given by s x) 2 + y) 2 = + ) 2 y x. x Thus we obtin x2 S =2π) yx) +y x)) 2 dx. x Thus we hve F x, y, y )=2πy +y ) 2 nd the Euler eqution 6..3) becomes [ ] d yy +y ) 2 ) /2 =. dx +y ) 2 ) /2 If we use 6..5) then we get To solve this eqution we let yy y ) 2 =. p = y so tht y = dp dx = dp dy dy dx = pdp dy nd this gives py dp dy = p2 +. This eqution is seprble, nd is integrted s follows pdp p 2 + = dy y 2 lnp2 + ) = lny)+c. Exponentiting both sides we get Solving for dy/dx we get y = α + ) ) 2 /2 dy. dx ) dy y 2 /2 dx = α, 6

7 nd seprting vribles gin, we get dy y/α)2 = dx. Now let y = α cosht) so tht y/α) 2 = cosh 2 t) = sinh 2 t), nd dy = α sinht) dt. sinht) dt dx = α = α dt = αt + c sinht) x b) so x = αt + β which implies t = α ) x β y = α cosh. α Thus the miniml surfce, if it exists, is cntenry. It remins to be seen whether the rbitrry constnts α nd β cn be chosen tht the curve psses through the points x,y ) nd x 2,y 2 ). This cn be done for ny two points in the upper hlf plne. In generl, the determine of the two constnts involves the solution os trnscendentl eqution which possess two, one or no solutions, depending on yx ) nd yx 2 ). Let us fix the initil point x,y ) to be, ). We show tht there re point which cnnot be reched to solve the extreml problem, i.e., the boundry vlue problem hs no soultion. First since we need y)=we hve β =αcosh α Let us denote λ = β/α so we hve ), or α = coshβ/α). coshx coshλ) λ) y = yx, λ) =. coshλ) Since y = sinhx coshλ) λ) we hve y ) = sinhλ). Thus s λ vries over ll rel numbers we get ll possible slopes. Now we try to find λ so tht the curve psses through the second point which we now denote s b, y b ). So we need to solve the eqution coshb coshλ) λ) yb, λ) = = y b. coshλ) We now show tht b, y b ) cn be chosen so tht this eqution hs no solution. First note tht cosht) = 2 et + e t ) t for ll t. hence coshb coshλ) λ) coshλ) b coshλ) λ) coshλ) = b λ coshλ) 7 b λ coshλ)

8 it is esy to see tht mx λ λ coshλ) = λ coshλ ) where λ is the positive root of the eqution Thus we hve nd finlly coshλ ) λ sinhλ )=. b yb, λ) b λ coshλ) b sinhλ ),, for ll λ. sinhλ ) So if b is chosen so tht b sinhλ ) = B> nd y b is then chosen so tht y b <B, then the BVP cnnot hve solution. Exmple 6.2. Let A nd B be two fixed points in the plne. The time it tkes for prticle to slide under the influence of grvity long curve pth) from A to B depends on the choice of the curve. The curve the requires the miniml time is clled brchistochrome. The problem of finding this curve is thus clled the brchistochrome Problem. The problem ws posed by John Bernoulli in 696. This problem plyed n importnt role in the development of the clculus of vritions. The problem ws solved by John Bernoulli, Jmes Bernoulli, Newton, nd L Hospitl. We will consider specil cse Fermt s Problem) tht results in solution tht is not so complicted. Nmely wnt to find curve y = yx) which minimizes the time decent for slidoing frictionless bed between y) = y > ) nd the finl point is yb) =. We ssume tht the speed of trvel in the prticulr medium is given t ech point by cx, y). Then ds cx, y) dt where ds is incrmentl rclength. Then the totl time of trvel is +y ) T = 2 dx. cx, y) Since the prticle is t rest t height y, nd since the potentil plus kenetic energies is constnt, the speed t height y is c = 2gy y) where g is the grvittionl constnt. Put nother wy, if s represents distnce on the curve y = yx) mesured from,y ), then ds dt = v where t represents time nd v is the velocity of the prticle. Since the motion, by ssumption, is frictionless with initil velocity v =nd is influenced only by grvity, the principle of conservtion of energy implies 2 mv2 = mgy y) 8

9 where y is the verticl distnce, nd m is the mss of the prticle. This implies v = 2gy. Hence ds dt = 2gy y) or dt = ds 2gy y) = +y ) 2 dx 2gy y). Therefore the time T tht the prticle tkes to get from,y ) to b, ) long the curve y = yx) is given by T = +y ) 2 dx. 2g y y) The problem is to find yx) to minimize T. In our nottion we hve F x, y, y +y )= ) 2 y y). Note tht F does not depend on x explicitly, so the Euler eqution hs the form y y F = c or y y)+y ) 2 ) ) /2 = c. From this we obtin first order ODE dy c 2 dx = y y) y y) with c n rbitrry constnt of integrtion. To solve this eqution we introduce prmetric substitution y = y + c 2 cost) )/2. It follows tht nd ) /2 y y = c 2 cost) )/2 ) dy c 2 dx = + c 2 /2 ) cost) )/2 c 2 /2 = + cost) )/2) c 2 cost))/2 c 2 cost))/2 ) /2 ) /2 cost)+)/2 + cost)) cost) 2 ) = = = cost))/2 cost)) cost)) 2 ) sin 2 /2 ) t) sint) = = cost)) 2 cost)) ) /2 9

10 Thus we obtin nd since we hve dx dt From this we obtin cost)) dy = sint) dt dy dx = sint) cost)) dy dt = c2 sinxt)) 2 cost)) c 2 = sint)) sint) 2 = c2 cost)). 2 x = 2 c2 t sint)), y = y + c 2 cost) )/ ) The constnt c must be chosen so tht y =when x = b. Set x = b nd y =in in 6..8), then solve ech of the equtions for c 2 c 2 = 2b t sint), c2 = Next we equte these expressions, eliminting c 2, to obtin or ht) b t sint) = y cost) ) cost) = t sint) 2y cost). y ). b As t vries from to +, the left hnd side of this lst expression, i.e., ht), vries from to. So for ny positive vlues y nd b we cn find vlue T so tht y ) ht )=. b From this we cn determine c by 2y c = cost ). In this wy we cn obtin the solution which describes cycloid curve prmeterized by {xt),yt)): t T }. The following numericl exmple corresponds to y =2nd b =.

11 2 y = 2 b =.5 y x We now give n exmple to show tht the extreml function my not be C. Exmple 6.3. Consider the problem of minimizing subject to the boundry conditions J = y 2 y ) 2 ) dx 6..9) y )=, y)=. It is obvious tht y =nd y = x both give miniml vlues for J, nmely J =. But it turns out tht neither of these stisfies the end conditions. In order to obtin solution tht stisifies these conditions we must build piecewise smooth solution. The function { for x< y = x for x is such solution. The solution stisfies the integrted form of the Euler-Lgrnge eqution nd ny smooth portion stisifes the differentil eqution form y y ) 2 + d [ y 2 y ) ] =. dx There re dditionl conditions tht sectionlly smooth extreml must stisfy for exmple, the Weierstrss-Erdmnn Corner Conditions. Theorem 6.. If y = y x) is piecewise C smooth reltive mximum or minimum) of J = fx, y, y ) dx

12 then t ll points x = c where y is discontinuous with jump discontinuity), the Weierstrss- Erdmnn Corner Conditions f y x=c = f y nd x=c + must be stisfied. y f y f) x=c =y f y f) x=c + Functionls of Severl Unknown Functions: Suppose we wnt to find the extreml of functionl of severl unknown functions J = F x, y,y,y 2,y 2,,y k,y k) dx. 6..) We proceed to derive Euler equtions by introducing functions η j nd ssume tht yj re extreml functions. Then we introduce the dmissible functions y j = yj + α j η j, j =, 2,,k where α j re sclr. We require tht J α j =, j =, 2,,k, for ll functions η j. αj = This gives us the system of k Euler-Lgrnge equtions d y j dx y j =, j =, 2,,k. Functionls with Higher Order Derivtives: Suppose we wish to minimize or mximize) functionl involving higher order derivtives. Consider, for exmple, J = Here we suppose tht the vlues of F x, y, dy ) dx, d2 y dx, dk y dx. 6..) 2 dx k y, dy dx, d2 y dx 2, dk ) y dx k ) re specified t x = nd x = b. Agin we introduce dmissible functions y = y + αη, 2

13 where y is n extreml nd η stisifes η) =ηb) = dη ) =dη dx dx b) = = dk ) η =dk ) dx ) η b) =. k ) dx k ) Repeting the usul procedure we obtin F y d dx F y + d2 dx F dk 2 y + + )k dx F k y k) =. Nturl Boundry Conditions: One other importnt considertion is tht for some problems we re not given fixed endpoint conditions. In this cse the procedure chnges some. Suppose we wnt to minimumize or mximize) J = F x, y, y ) dx but without specifying vlues of y t x = or x =. In this cse the difference, αη of the minimimum y nd the vried function y + αη need not be zero t the ends since no end conditions re required of y). However, the integrl term in 6..2) must still vnish, since mong ll possible dmissible functions η re those which vnish t the end points nd we cn pply Lemm 6. to obtin the Euler eqution. Now since the fisrt term is zero we now find tht [ η] [ η] = y x= y x= for ll dmissible η, which in turn mens for ll vlues of η) nd η) if y is not prescribed t either end. In this cse we hve the dditionl requirements [ ] =, y x= [ ] = 6..2) y x= Further, if y) is not prescribed then the first identity in 6..2) must hold nd if y) is not prescribed then the second identity in 6..2) must hold. The conditions in 6..2) re clled the nturl boundry conditions. In some cses it my lso hppen tht the integrnd F or nd/or d/dx) y re discontin- y uous t one or more points inside the intervl, ) but the conditions ssumed bove re stisifed in ech subintervl seprted by these points. For exmple, suppose there is one point of discontinuity x = c. Then the integrl 6..) must be expressed s two integrls, one over,c ) nd the other over c +, ): J α = α c F x, y + αη, y + αη ) dx + ) F x, y + αη, y + αη ) dx c ) 3

14 We obtin c = y d ) ηdx+ dx y c y d ) ηdx dx y + [ ] c [ ] ) + y η y η. c + If we require tht the minimizing function y be continuous on, ) in prticulr t x = c), then then we lso need ll dmissible functions y + αη to lso be continuous. It follows tht ηc )=ηc + )=ηc). From this we cn write c = y d ) ηdx+ dx y c y d ) ηdx dx y + [ ] [ ] [ ] [ ] ) c + η) η) ηc). y y y c y + Now since η), η) nd ηc) re rbitrry we cn conclude tht Euler eqution y d ) = dx y must hold in ech subintervl,c ) nd c +, ). We lso see tht must vnish t ny endpoint y for which y hs not been prescribed. Finlly, the nturl trnsition conditions must hold: yc + )=yc ), lim = lim x c + y x c y. 6..4) These conditions require tht y be continuous but tht y my be discontinuous t x = c. Exmple 6.4. Consider finding the sttionry J = T y ) 2 ρω 2 y 2) dx 6..5) where T, ρ, nd ω re given constnts or functions of x). The Euler eqution becomes d T dy ) + ρω 2 y =, dx dx regrdless of wht is presecribed t the endpoints of the intervl. If, for exmple, T, ρ nd ω re positive constnts, then the extremls re of the form ) y = c cosαx)+c 2 sinαx), α 2 = ρω2. T 4

15 . If y)=nd y) = re prescribed then c =, nd c 2 sinα) =which implies y = sinαx) for α π, 2π,. sinα) 2. If y)=is prescribed but y) is not prescribed then from [ ] ) [T y ) 2 + ρω 2 y 2 ] = =, y y x= we obtin Ty )= nd hence y = cosαx) for α π cosα) 2, 3π 2,. 3. If neither y) or y) is prescribed then the conditions require Ty ) =, Ty )= nd we otin { α π, 2π, ) y = c cosαx) α = π, 2π, ), where c is n rbitrry constnt. In the exceptionl cses bove there re no sttionry functions. The limiting cse α =must be treted s specil cse where we replce sinαx) by sinαx)/α then pss to the limit. Let us now consider the cse in which T nd ρ re step functions with constnt vlues on subintervls,c) nd c, ) with <c<: T x) = { T x<c T 2 c<x, ρx) = { ρ x<c ρ 2 c<x. We ssume tht T, T 2, ρ, ρ 2, ω re positive constnts nd we ssume tht y)=, y)= hve been prescribed. The we obtin from the Euler equtions on the subintervls: c cosα x)+c 2 sinα x) x<c y =, d cosα 2 x)+d 2 sinα 2 x) c<x< where c, c 2, d, d 2 re rbitrry constnts nd αj 2 = ρ jω 2, j =, 2. T j The nturl trnsition conditions give yc + )=yc ), T 2 y c + )=T y c ). In this cse we obtin system of four equtions in four unknowns which cn be solved for except for set of exceptionl vlues of α. 5

16 Severl Independent Vribles: In the cse of functions of more thn one vrible we might encounter n extreml problem of the form J = F x,x 2,,x k,y, y, y,, y ) dx dx dx k, 6..6) x x 2 x k D where D is region in R k. The usul rguments considered in more detil in the 2-dimensionl cse below led to the following. We ssume tht η is function of x,x 2,,x k ) tht vnishes on the boundry of D. We replce y by y + αη nd compute the derivtive with respect to α to obtin D y + k j= ) η dx dx dx k =, z j x j where z j = y x j. To use integrtion by prts in higher dimensions we recll the divergence theorem which sttes Theorem 6.2. For vector vlued function f div f) dv = D D f nds 6..7) where dv is volume mesure, ds is surfce mesure, n is the outwrd norml vector to the boundry of D which is denoted by D. With this we hve k x j D j= η ) [ dv = η,, ] ) n ds, z j D z z k so tht lso using the product rule) k ) η dv = η z j x j D j= D [,, ] ) n ds η z z k D k j= x j ) dv, z j Now we impose the boundry condition η =on D. Thus we conclude y k j= x j ) z j =, z j = y x j. 6..8) If the function y is not prescribed priori on D, then we cnnot ssume tht η =on D. Insted we hve n =, z j on the prt of the boundry D where y is not prescribed. 6 z j = y x j,

17 Remrk 6.2. It is importnt tht we point out tht in this whole discussion we hve only considered necessry conditions for n extrem. Tht is n extrem will stisfy the Euler-lgrnge eqution. Recll tht from clculus necessry condition for mimimum or mximum) t x = x is tht f x )=but just from this we cnnot conclude tht x gives locl extrem for exmple, it could give point of inflection). On the other hnd the second derivtive test when it is pplicble) does give us sufficient conditions for n extrem f x ) > implies tht x is minimum nd f x ) < implies tht x is mximum). This is clled the second derivtive test. In the clculus of vritions there is n nlogous sufficient condition tht comes from the so-clled second vrition. Actully this second vrition is seldom used in prctice. Generlly other methods re used to verify tht sttionry vlue is ctully n extrem. These rguments re generlly somewht more complicted nd won t be considered here. Exmple 6.5 Plteu s Problem). As n exmple let us consider the problem of finding surfce pssing through given closed curve C in spce nd hving minimum surfce re bounded by C. If we ssume tht the surfce cn be represented s function z = zx, y) then the re to be minimized is given by the integrl J = R ) +z 2 x + zy 2 /2 dxdy 6..9) where R, is the region in the xy plne bounded by the projection C of C onto the xy plne nd where z is given long C, i.e., if xt),yt)) is prmeteriztion of C, then the vlues of z on C re prescribed in terms of further prmeteriztion of C given by zxt),yt)) = ht) where we re given h. Wefind F x, y, z, z x,z y )= ) +zx 2 + zy 2 /2 nd the Euler eqution is [ ] [ ] z x ) x +z 2 x + zy 2 /2 + z y ) y +z 2 x + zy 2 /2 =. After some simplifiction this cn be written s Note this is prtil differentil eqution. + z 2 y)z xx 2z x z y z xy ++z 2 x)z yy =. Exmple 6.6 Dirichlet Problem). Next we consider n exmple tht rises in tremendous number of prcticl situtions the Dirichlet Problem. In two dimensions we must minimize the functionl D[u] = R ) u 2 x + u 2 y dxdy 6..2) where R, is the region in the xy plne bounded by the projection C of C onto the xy plne nd where u is given long C : uxt),yt)) = gt), t [, b]. 7

18 In this cse find F x, y, u, u x,u y )=u 2 x + u 2 y nd the Euler eqution is u xx + u yy =. This eqution is clled Lplce s eqution. Problems with Contrints: Lgrnge Multipliers: In mny problems we not only need to minimize functionl but this must be crried out subject to some dditionl constrints. An importnt method for deling with problems of this type is the Method of Lgrnge Multipliers. Let us suppose tht we wnt to minimize functionl J[y] = F x, y, y ) dx 6..2) where y is prescribed t the endpoints by y) = y nd y) = y nd y is subject to the extr constrint tht K = Gx, y, y ) dx = l ) We consider vrition of y in the form y + α η + α 2 η 2 where η nd η 2 re continuously differentible nd vnish t the endpoints. This we consider Jα,α 2 ) F x, y + α η + α 2 η 2,y + α η + α 2 η 2) dx Kα,α 2 ) Gx, y + α η + α 2 η 2,y + α η + α 2 η 2), dx nd it must be true tht Jα,α 2 ) tkes on minimum vlue subject to the constrint Kα,α 2 )= when α = α 2 =. Thus s in multidimensionl clculus we wnt to find minimum of function of α,α 2 ) subject to constrint. One of the most useful tools for solving problems of this type is the method of Lgrnge Multipliers. We consider [Jα,α 2 )+λkα,α 2 )] α α =α 2 =, 6..23) = [Jα,α 2 )+λkα,α 2 )] α α =α 2 = ) = 2 If we crry out the necessry steps nd do integrtion by prts in ech of the pprorite terms, we end up with the two equtions [ y d )) dx y G + λ y d dx [ y d )) G + λ dx y y d dx 8 G y G y ))] η dx =, 6..25) ))] η 2 dx = )

19 Generlly it cn be ssumed tht the coefficient of λη 2 in 6..26) is not identiclly zero so it cn be ssumed tht G y d )) G η 2 dx. dx y x In this cse we cn choose λ so tht 6..26) is stisfied. Then since η in 6..25) is rbitrry its cosefficient in 6..25) must vnish. We obtin ) d F + λg) F + λg) = ) y dx y The generl solution of this problem will involve the constnt prmeter λ s well s the two constnts of integrtion. This corresponds to the fct tht we hve two end conditions nd one extr constrint given in 6..22). Remrk 6.3. In order to mximize or minimize) n integrl b Fdxsubject to constrint Gdx = l, first write H = F + λg, where λ is constnt, nd mximize or minimize) H dx subject to no contrint. Crry the Lgrnge multiplier λ through the clcultion, nd determine it, together with the constnts of integrtion rising in the solution of Euler s eqution, so tht the constrint is stisfied nd so tht the end conditions re stisfied. b Gdx= l When one of the end conditions y) = y or y) = y ) is not imposed then the condition H =must be substituted for it t the end in question. y Exmple 6.7. Consider the problem of determining the curve of fixed length l which psses through the points, ) nd, ) nd for which the re between the curve nd the x xis is mximum. We re thus led to mximize the integrl J[y] = ydx 6..28) subject to the end constrints y)=nd y)=nd lso subject to the constrint K = The corresponding Euler eqution for +y ) 2) /2 dx = l. H = y + λ +y ) 2) /2 is [ ] λ d y dx +y ) 2) /2 =. 9

20 After integrion nd simplifiction y ) 2 = Solving for y nd integrting gin we find x c ) 2 [λ 2 x c ) 2 ]. y = ± [ λ 2 x c ) 2] /2 + c2. Rerrnging terms we find s you would expect) x c ) 2 +y c 2 ) 2 = λ 2 corresponding to rcs of circle. The constnts c, c 2 nd λ re to be determined so tht the rc psses through, ),, ) nd hs length l. Exmple 6.8 Sturm-Liouville Problems). The problem of finding the eigenvlues nd eigenvectors of Sturm-Liouville problem cn be formulted s vritionl problem. Consider the functionl J[y] = px)y x)) 2 + qx)yx) 2) dx 6..29) subject to constrint K[y] = ωx)yx) 2 dx =, 6..3) where the functions p, q nd ω re positive on the intervl x. We propose to minimize 6..29) subject to 6..3), which using Lgrnge multiplyiers is equivlent to minimizing H[y] =J[y]+λK[y]. The Euler-Lgrnge eqution is px)y ) qx)y + λωx)y =, subject to ny set of boundry conditions for which px)y η =, where ηx) is ny dmissible function. If y is the minimizer of J subject to K =, then λ is n eigenvlue with eigenfunction y. For exmple, if we hve specified y) = y nd y) = y, then we need η) = η)=. If η is not specified t nd then we must hve the ntrul boundry conditions y ) = nd y ) =. These correspond to Neumnn nd Dirichlet boundry conditions. For more generl boundry conditions we must replce J with J[y] = px)y x)) 2 + qx)yx) 2) dx + σ)p)y 2 ) + σ)p)y 2 ) 6..3) 2

21 still subject to 6..3). The Euler-Lgrnge eqution is gin but now the nturl boundry conditions re px)y ) qx)y + λωx)y =, y ) + σ)y)=, y ) + σ)y)=. The conclusions in Exmple 6.8) led to n importnt point in the history of mthemtics. Minimum Principle. If y renders J[y] miniml subject to the restriction K[y] =, then px)y ) qx)y + λωx)y =nd the miniml vlue J[y] =λ. 2. Suppose the first k eigenfunctions y,y 2,,y k of px)y ) qx)y + λωx)y = with λ λ 2 λ k known. Then the k +)st eigenpir y k+),λ k+) isdetermined s the solution of the minimum of J[y] miniml subject to the restriction K[y] =nd the k dditionl constrints yx)y j x) dx =, j =, 2,,k. Furthermore we hve is the k +)st eigenvlue. J[y k+) ]=λ k+) λ k, Assignment Chpter 6. Consider finding continuously differentible function yx) which minimizes J[y] = +y ) 2 ) dx, y)=, y)=. ) Obtin the relevnt Euler eqution, nd show tht the sttionry function is y = x. b) With yx) =x nd the specil choice ηx) =x x), clculte Jα) =J[y + αη] = nd verify directly tht dj dα α) α= =. F x, yx)+αηx),y x)+αη x)) dx 2

22 c) By writing yx) =x + ux), show tht the problem becomes J =2+ u x)) 2 dx = minimum, where u) = u) =. Use this result to deduce tht yx) = x is the required minimum. 2. Find extreml curves for J = 3. Find extreml curves for J = y 2 y ) 2 2y cosx)) dx y ) 2 x 3 dx 4. Find the extrem for J = 2 +y ) 2 dx, x y)=, y2)=. 5. Find the extrem for J = x y) 2 dx. 6. Find the rc with the shortest distnce between the two points,) nd,b). Here you cn ssume tht the solution cn be expressed s function of one vrible y = yx). 7. Find the curve yx), x,withy) = y)=nd fixed rclength 3π/4 tht gives mximum re A = yx) dx. Note tht this is the Exmple 6.7. Thus this problem mounts to filling in ll the detils i.e., solving the differentil eqution, finding the constnts, etc.) for Exmple 6.7 with specific l. ) 8. Which curve minimizes 2 y ) 2 + yy + y + y dx when y) nd y) re not specified. 9. Determine the sttionry functions ssocited with the integrl where α nd β re constnts: ) y)=nd y)=, b) y)= c) y)= d) No end conditions. Note ny exceptionl cses. J = y ) 2 2αyy 2βy ) dx, 22

23 . Find the sttionry solutions for the vritionl problem with constrint: Minimize subject to the constrint References J[y] = π y ) 2 dx, y)=, yπ) =, π y 2 dx =. [] G.A. Bliss, Lectures on the Clculus of Vritions, Unmiv. of chicgo Press, 946. [2] G.M. Ewing, Clculus of Vritions with Applictions, Dover, 985. [3] I.M. Gelfnd, S.V, Fomin, Clculus of Vritions, Prentice Hll, 963. [4] F.B. Hildebrnd, Methods of Applied mthemtics, Dover, 992. [5] J. Keener, Principles of Applied mthemtics, Addison-Wesley, 988. [6] H. Sgn, Introduction to the clculus of vritions, Dover, 992. [7] D.R. Smith, Vritionl Methods in Optimiztion, Dover, 998. [8] I. Stkgold, Green s Functions nd Boundry Vlue Problems, Wiley-Interscience, 979. [9] E. Zeidler, Nonliner Functionl Anlysis nd its Applictions III: Vrintionl Methods nd Optimiztion, Springer-Verlg,