Chpter 4 Additionl Vritionl Concepts 137 In the previous chpter we considered clculus o vrition problems which hd ixed boundry conditions. Tht is, in one dimension the end point conditions were speciied. In two nd three dimensions, boundry condition or surce condition ws speciied. In this chpter we shll consider other types o boundry conditions together with subsidiry conditions or constrints imposed upon the clss o dmissible solutions which produce n extreml vlue or given unctionl. We shll lso exmine necessry conditions or mximum or minimum vlue o unctionl to exist. Let us begin by exmining the onedimensionl cse where either one or both boundry conditions re not prescribed. Nturl boundry conditions Consider the problem o inding unction y = y(x) such tht the unctionl I = I(y) = (x, y, y ) (4.1) hs sttionry vlue where either one or both o the end point conditions y( )=y 1 nd y(x 2 )=y 2 re not prescribed. One proceeds exctly s we hve done previously. Assume tht y = y(x) produces n extreml vlue nd construct the clss o comprison unctions Y (x) =y(x) +ɛη(x). Substitute Y (x) into the unctionl given by eqution (4.1) to obtin I = I(ɛ) = (x, y + ɛη, y + ɛη ). By hypothesis, the unctionl I hs sttionry vlue t ɛ =0so tht di x2 ( = I (0) = dɛ ɛ=0 η + ) η =0. (4.2) Now integrte the second term in eqution (4.2) using integrtion by prts to obtin di = I (0) = dɛ ɛ=0 η x2 x2 [ + d ( )] η=0. (4.3) Assume tht the Euler-Lgrnge eqution d ( ) =0 (4.4) is stisied over the intervl x x 2, then in order or eqution (4.3) to be stisied it is necessry tht the irst term o eqution (4.3) lso equl zero. This requires tht η x2 = x=x 2 η(x 2 ) x= η( )=0. (4.5)
138 In the cse o ixed end points, the Euler-Lgrnge eqution together with the end conditions η( )=0nd η(x 2 )=0gurntees tht the eqution (4.5) is stisied. In the cse both end points re vrible, then the terms η( ) nd η(x 2 ) re rbitrry, nd so in order or eqution (4.5) to be stisied one must require in ddition to the Euler-Lgrnge eqution, the end point conditions x= =0, nd x=x 2 =0. (4.6) These conditions re clled the nturl boundry conditions or trnsverslity conditions ssocited with the extremum problem. The cse o mixed boundry conditions occurs when one end o the curve is ixed nd the other end cn vry. The bove considertions give rise to the ollowing boundry cses. Cse 1: (Fixed end points) I y( )=y 1 nd y(x 2 )=y 2 re prescribed, then there is no vrition o the end points so tht η( )=0nd η( )=0. Cse 2: (Mixed end condition) I y( )=y 1 is given nd y(x 2 ) is unknown, then one must impose the nturl boundry condition =0nd η(x 1 )=0. x=x 2 Cse 3: (Mixed end condition) I y(x 2 )=y 2 is given nd y( ) is unknown, then one must impose the nturl boundry condition =0nd η(x 2 )=0. x= Cse 4: (Vrible end points) I y = y(x) is not prescribed t the end points o the given intervl x x 2, then the prtil derivtive must stisy the nturl boundry conditions given by equtions (4.6). Note tht the requirement tht η =0t n end point is equivlent to speciying the vlue o the dependent vrible y t n end point. Boundry conditions where η =0or δy =0, where the vlues o y re speciied t boundry, re clled essentil boundry conditions sometimes reerred to s Dirichlet boundry conditions or geometric boundry conditions. Nturl boundry conditions re sometimes reerred to s Neumnn boundry conditions or dynmic boundry conditions. Mixed boundry vlue problems, sometimes reerred to s Robin boundry vlue problems, occur when both essentil nd nturl boundry re speciied on portions o the boundry. A generl rule is tht the vnishing o the vrition η on boundry is n essentil boundry condition nd the vnishing o the term tht multiplies η is clled nturl boundry condition. This generl rule, nd the bove terminology, with slight modiictions, cn be pplied to one, two nd three dimensionl problems. Exmple 4-1. Nturl boundry condition Find the curve y = y(x) producing the shortest distnce between the points x 0 nd subject to nturl boundry conditions. Solution: Let ds 2 = 2 + dy 2 denote n element o distnce squred in Crtesin coordintes. One cn then orm the unctionl I = x1 x 0 ds = x1 x 0 1+ ( ) 2 dy.
139 The integrnd or this unctionl is = 1+(y ) 2 with prtil derivtives =0, = y 1+(y ). 2 The Euler-Lgrnge eqution ssocited with this unctionl is y =0which is subject to the nturl boundry conditions y (x 0 ) = x=x 0 1+(y (x 0 )) =0 2 y ( ) = x= 1+(y ( )) =0 2 The Euler-Lgrnge eqution hs the solution y = y(x) =Ax+B where A, B re constnts. The nturl boundry conditions require tht A =0nd so the solution is y = B = constnt. Nturl boundry conditions or other unctionls Consider the unctionl b I = (x, y, y,y ) (4.7) where we hold the vrition in x constnt nd consider only the vrition in y. This unctionl hs irst vrition b ( δi = δy + δy + ) δy =0. (4.8) Use integrtion by prts on the second term o eqution (4.8) using U =, du = d ( ), dv = δy, V = δy nd then use integrtion by prts on the third term in eqution (4.8) using U =, du = d ( ), dv = δy, V = δy nd veriy tht eqution (4.8) cn be expressed in the orm δi = b [ δy d ( ) δy d ( ) ] δy + δy b + δy b =0 (4.9) Now use integrtion by prts gin on the third term under the integrl in eqution (4.9) to show b [ δi = d ( ) ( )] [ + d2 δy + 2 d ( )] δy b + b δy =0. (4.10) The Euler-Lgrnge eqution ssocited with the unctionl given by eqution (4.7) is d ( ) ( ) + d2 =0. 2
140 This is ourth order ordinry dierentil eqution so tht the generl solution will contin our rbitrry constnts. The boundry conditions come rom the requirement tht the remining terms on the right-hnd side o eqution (4.10) must equl zero. This requires [ d ( )] δy b =0 nd δy b =0. (4.11) This produces the essentil boundry conditions δy nd δy equl to zero t x = nd x = b producing the conditions tht y(),y(b) nd y (),y (b) re speciied t the end points. ( ) Alterntively, one cn use the nturl boundry conditions tht d nd re zero t the end points x = nd x = b, or one cn employ some combintion o mixed boundry conditions determined rom the eqution (4.11). For exmple, one cn select δy =0t x = nd x = b together with =0nd x = nd x = b s mixed boundry condition involving both essentil nd nturl boundry conditions. The other combintion is to ssume δy =0 ( ) t x = nd x = b together with d = t the end points. Note tht the conditions δy =0nd δy =0implies tht y(),y(b),y (),y (b) re ll speciied so tht there is no vrition o these quntities t the end points. Exmple 4-2. Nturl boundry condition Find the unction y = y(x) such tht the unctionl I = b [(x)(y ) 2 b(x)(y ) 2 + c(x)y 2 ] is n extremum. Solution: Here = (x)(y ) 2 b(x)(y ) 2 + c(x)y 2 with prtil derivtives =2c(x)y, = 2b(x)y, =2(x)y The Euler-Lgrnge eqution ssocited with the bove unctionl is the ourth order ordinry dierentil eqution c(x)y + d (b(x)y )+ d2 2 ((x)y )=0, x b Vrious types o boundry conditions cn be pplied to obtin solution to this dierentil eqution. Cse 1 I δy =0nd δy =0t the end points, then y(),y (),y(b),y (b) must be prescribed. Cse 2 I δy =0nd δy = 0t the end points, then y() nd y(b) must be prescribed nd the nturl boundry conditions x= =0, nd x=b =0 must be stisied. This implies y () nd y (b) hve prescribed vlues.
141 Cse 3 I δy = 0nd δy =0t the end points, then y () nd y (b) re prescribed nd the nturl boundry conditions [ d ( )] [ =0, nd x= d ( )] =0, x=b must be stisied. Cse 4 I δy = 0nd δy = 0t the end points, then the nturl boundry conditions =0 nd d ( ) =0 t the end points x = nd x = b must be stisied. More nturl boundry conditions Consider the unctionl I = R (x,y,w,w x,w y ) dy (4.12) where w = w(x, y) is unction o x nd y nd R is region enclosed by simple closed curve C = R. We use the vritionl nottion nd require tht t n extremum the irst vrition is zero so tht [ δi = δw + δw x + ] δw y dy =0. R w w x w y (4.13) Now use the Green s theorem in the plne ( N x M ).. dy =. M+ Ndy (4.14) R with N = w x δw nd M = w y δw to integrte the lst two terms under the integrl on the right-hnd side o eqution (4.13). Note tht or the bove choices or M nd N we hve N x = w x δw x + x ( w x ) δw nd C M = w y δw y ( nd so the use o the Green s theorem produces the result ( δw x + ) (.. δw y dy =. δw + ) δw dy R w x w y C w y w x [ ( ) δw + ( ) ] δw dy R x w x w y w y ) δw (4.15) Substitute the result rom eqution (4.15) into the eqution (4.13) nd simpliy to obtin [ δi = R w ( ) ( )] δw dy x w x w y (.. +. δw + ) (4.16) dy =0. w y w x C I the eqution (4.16) is to be zero, then w = w(x, y) must stisy the Euler-Lgrnge eqution w ( ) ( ) =0 x w x w y
142 nd the boundry conditions come rom n nlysis o the remining line integrl term. I δw =0, then the boundry conditions required re or the unction w = w(x, y) to be speciied everywhere long the boundry curve C = R. I δw = 0, then we require tht w y + w x dy This condition cn be written in terms o the unit norml vector to the boundry curve C = R. Note tht i r = x(s) ê 1 +y(s) ê 2 is the position vector deining the boundry curve C, then d r ds = ˆt = ds ê1 + dy ds ê2 is the unit tngent vector to point on the boundry curve. The cross product (x,y) R ê 1 ê 2 ê 3 ˆn = ˆt ê 3 = dy 0 ds ds 0 0 1 = dy ds ê1 gives the unit norml vector to the boundry curve. eqution (4.17) cn then be written in the orm ( ˆn ê 1 + ) ê 2 w x w y (x,y) R =0. (4.17) ds ê2 The boundry condition given by =0 (4.18) where ˆn = dy ds ê1 ds ê2 is unit norml vector to the boundry curve C = R. The condition (4.18), or (4.17), is clled the nturl boundry condition ssocited with the unctionl given by eqution (4.12). Tests or mxim nd minim So r we hve been solving the Euler-Lgrnge eqution nd stting tht either mximum or minimum vlue exists. We hve not ctully proved these results. The ollowing re two tests or extremum problems ssocited with the unctionl given by eqution (4.1). These tests re known s Legendre s test nd Jcobi s test. The Legendre nd Jcobi nlysis Consider the unctionl I = I(y) = (x, y, y ) (4.19) nd ssume tht y = y(x) produces n extremum. I we replce y by the set o comprison unctions Y = y + ɛη, where we ssume wek vritions with η( )=0nd η(x 2 )=0. One cn then tret I s unction o ɛ nd write I = I(ɛ) which cn be expnded in Tylor series bout ɛ =0. This produces the result I = I(ɛ) = (x, y + ɛη, y + ɛη ) = I(0) + I (0)ɛ + I (0) ɛ2 2! + I (0) ɛ3 + (4.20) 3!
143 The chnge in the vlue o the unctionl I is given by I = [(x, y + ɛη, y + ɛη ) (x, y, y )] (4.21) Expnded the integrnd o eqution (4.21) in Tylor series bout ɛ =0to produced (x, y + ɛη, y + ɛη ) (x, y, y )= ɛ (x, y + ɛη, y + ɛη ) + ɛ2 2 (x, y + ɛη, y + ɛη ) 1! ɛ ɛ=0 2! ɛ 2 + ɛ=0 + ɛn 1 (n 1) (x, y + ɛη, y + ɛη ) (n 1)! ɛ n 1 + R n ɛ=0 (4.22) An integrtion o both sides o eqution (4.22) rom to x 2 produces I = ɛ 1! ɛ ɛ=0 (x, y + ɛη, y + ɛη ) + ɛn 1 x2 (n 1)! + ɛ2 2! (n 1) (x, y + ɛη, y + ɛη ) ɛ (n 1) ɛ=0 ɛ 2 ɛ=0 2 (x, y + ɛη, y + ɛη ) + R n + (4.23) The coeicients o quntities ɛ/1!,ɛ 2 /2!,...,ɛ (n 1) /(n 1)! re reerred to s the irst, second,..., (n-1)st vritions o the unctionl I nd written s δi 1, δi 2,...δI (n 1). One cn clculte the derivtives in eqution (4.23) nd veriy tht these vritions re given by [ δi 1 = η + ] η (4.24) [ 2 δi 2 = 2 η2 +2 2 ηη + 2 ] 2 (η ) 2 (4.25) [ 3 ] δi 3 = 3 η3 +3 3 2 η2 η +3 3 2 ηη 2 + 3 η 3 (4.26) 3. Then the chnge in the vlue o the unctionl I cn be written I = ɛ 1! δi 1 + ɛ2 2! δi 2 + ɛ3 3! δi 3 + + ɛ(n 1) x2 (n 1)! δi n 1 + R n (4.27) Some textbooks write these vritions using the nottions The reltion between these dierent nottions is I = δi + 1 2! δ2 I + 1 3! δ3 I + (4.28) δ m I = ɛ m δi m = I (m) (0)ɛ m or m =1, 2, 3,... (4.29) Exmine the irst vrition nd ssume tht η =0t the end points o the integrtion intervl. Use integrtion by prts on the second term in eqution (4.24) to obtin [ ] x2 x2 [ δi 1 = η + d ( )] η. (4.30)
144 The end point conditions insures tht the irst term in eqution (4.30) is zero. In order or the second term in eqution (4.30) to be zero the unction y = y(x) must stisy the Euler- Lgrnge eqution subject to boundry conditions y( )=y 1 nd y(x 2 )=y 2. I these conditions re stisied one cn write tht di dɛ =0is necessry condition tht the unctionl ssume ɛ=0 sttionry vlue. We ssume tht the extreml curves y which re solutions o the Euler-Lgrnge equtions nd the derivtives o y re continuous unctions. However, this is not lwys the cse. Discontinuous solutions to the Euler-Lgrnge eqution re lso possible. In these cses the unction y must be continuous but the irst or higher derivtives my be discontinuous unctions o the independent vrible x. Discontinuous solutions re represented by clculting continuous curves over dierent sub-intervls o the solution domin nd then requiring continuity o these solutions t the end points o the sub-intervls. In this wy continuous curve is constructed by joining the solution pieces t the end points o the sub-intervls so tht the resulting curve is continuous. In this type o construction one is conronted with corners t the junction points where the let nd right-hnd derivtives re not lwys the sme. At ll corner points the quntities nd y must be continuous. These conditions re ssocited with the Weirstrss-Erdmnn corner conditions to be discussed lter in this chpter. For present we ssume continuity or the unctions y nd its derivtive dy, where y is solution o the Euler-Lgrnge eqution. I the irst vrition is zero, δi 1 =0, then the sign o the unctionl chnge I is determined by the sign o δi 2. I δi 2 > 0 is positive nd o constnt sign t ll points o the extreml rc y(x), then reltive minimum exists. I δi 2 < 0, is negtive nd o constnt sign t ll points o the extreml rc y(x), then reltive mximum exists. Note tht the mximum or minimum vlue ssigned to the unctionl must be independent o the vlue o η nd ɛ. I δi 2 chnges sign over the extreml rc, then the sttionry vlue is neither mximum or minimum. Exmine eqution (4.27) nd note tht i both δi 1 nd δi 2 re zero, then the sign o I depends upon the sign o ɛ 3 δi 3 which chnges sign or ɛ>0 nd ɛ<0, so tht there cn be no mximum or minimum vlue unless δi 3 is lso zero. Consequently, or these conditions the sign o I is determined by n exmintion o the sign o the δi 4 vrition, i it is dierent rom zero. We wish to exmine the second vrition [ 2 δi 2 = 2 η2 +2 2 ηη + 2 ] 2 (η ) 2 (4.31) under the conditions o wek vrition where η( )=0, η(x 2 )=0nd the Euler-Lgrnge eqution y d ( y )=0is stisied by unction y(x) which stisies end point conditions y( ) = y 1 nd y(x 2 ) = y 2. I y(x) is solution o the Euler-Lgrnge eqution nd the unctionl I hs minimum vlue, then δi 2 must be positive or ll vlues o η over the intervl x x 2. I y(x) is solution o the Euler-Lgrnge eqution nd the unctionl I hs mximum vlue, then δi 2 must be negtive or ll vlues o η over the intervl x x 2.
145 The second vrition cn be written in severl dierent orms or nlysis. One orm is to integrte the middle term in eqution (4.31) using integrtion by prts to obtin [ δi 2 = η 2 2 ] x2 + { [ η 2 2 d ( 2 )] [ +(η ) 2 2 ]} (4.32) 2 2 Assume η is zero t the end points, then one cn sy tht necessry nd suicient condition or I to be minimum is or δi 2 > 0 which requires tht 2 2 d ( 2 ) > 0 nd or ll smooth wek vritions η. One cn lso write the second vrition in the orm δi 2 = 2 2 > 0 (4.33) {[ ] η 2 yy + ηη yy + η [η yy + η y y ]} (4.34) where subscripts denote prtil dierentition. Integrte the second hl o the integrl (4.34) using integrtion by prts to obtin ter simpliiction δi 2 = [ η 2 yy + ηη ] x2 [ x2 y y + η 2 x yy η 2 d 1 ( yy ) η d ] (η y y ) (4.35) By ssumption the irst term is zero t the end points. We now express the eqution (4.35) using dierentil opertor L( )nd write δi 2 = ηl(η) (4.36) where L(η) is the dierentil opertor L(η) = d [ ] [ dη y y yy d ] ( yy ) η (4.37) The dierentil opertor (4.37) hs the bsic orm L(u) = d [ p(x) du ] q(x)u (4.38) where p(x) = y y nd q(x) = yy d ( yy ) (4.39) nd the dierentil eqution L(u) = d [ p(x) du ] q(x)u =0, x x 2 (4.40) is known s Jcobi s dierentil eqution. The opertor L( )is sel-djoint Sturm-Liouville opertor which stisies the Lgrnge identity ul(η) ηl(u) = d [p(x)(uη ηu )] (4.41)
146 We mke use o the Lgrnge identity nd write the second vrition (4.36) in the orm η x2 ( η δi 2 = ul(η) = ηl(u)+ d ) u u [p(x)(uη ηu ] (4.42) This orm cn be simpliied by using integrtion by prts on the lst integrl with U =η/u ( uη ηu ) du = The second vrition cn now be expressed in the orm u 2 dv = d [p(x)(uη ηu )] V =p(x)(uη ηu ) (4.43) η 2 δi 2 = u L(u) η u p(x)(uη ηu ) x2 x2 ( uη ηu ) 2 + p(x) u We substitute the vlue p(x) = y y second vrition s δi 2 = rom eqution (4.39) nd rerrnge terms to write the η 2 u L(u) + η u y y (ηu uη ) x2 x2 + y y ) 2 (η η u (4.44) u Let us nlyze the second vrition given by eqution (4.44). For the time being we shll ssume tht the ollowing conditions re stisied. Condition (i) Assume u is nonzero solution to the Jcobi dierentil eqution L(u) =0or x x 2, with initil conditions u( )=0nd u ( ) = 0. Condition (ii) Assume the end point conditions η( )=0nd η(x 2 )=0so tht we re ssured tht the middle term in eqution (4.44) is zero. Condition (iii) Assume ( ) 2 η η u u = 0so tht the sign o the third term in eqution (4.44) is determined by the sign o y y. then one cn sy tht The condition y y > 0 or x x 2 is necessry condition or the unctionl I(y) to be minimum. The condition y y < 0 or x x 2 is necessry condition or the unctionl I(y) to be mximum. These re known s Legendre s necessry conditions or n extreml to exist. We shll return to nlyze the bove conditions ter irst hving developed some necessry bckground mteril ssocited with the Jcobi dierentil eqution. Bckground mteril or the Jcobi dierentil eqution We ssume tht y = y(x, c 1,c 2 ) is two-prmeter solution o the Euler-Lgrnge eqution (x, y, y ) d ( (x, y, y ) ) =0, x 1 x x 2 (4.45)