Lecture 10: Euler s Equations for Multivariable

Transcription

1 Lecture 0: Euler s Equatons for Multvarable Problems Let s say we re tryng to mnmze an ntegral of the form: {,,,,,, ; } J f y y y y y y d We can start by wrtng each of the y s as we dd before: y (, ) ( 0) ( ) α y + αη Remember that y (0) s the functon we want J f { y, y, y, y,, y, y ; } d

2 Applyng the chan rule, we fnd: J f y f y f y f y d d α + y α y α + + y y f y f y d y y Whch means we have ntegrals of the type encountered n the one-varable problem J So, for to be 0, each ntegral must be zero Ths means we have to satsfy Euler equatons, of the form: f d f y d y 0

3 Euler s Equaton wth Constrants In some cases, we want to restrct the set of paths one can take between ponts and, and fnd the mnmum of J subect to ths constrant For eample, we mght want to fnd the shortest dstance between two ponts on a surface wth a fed shape If ths happens, our dervaton of the many-varable Euler Equaton s altered. We stll have: J f y f y f y f y d d α + y α y α + + y y f y f y d 0 when α 0 y y but now the y are not ndependent, so we can t assume that each ntegral s zero ndvdually

4 Usng ntegraton by parts on each ntegral gves: J f y d f y f y d f y d d α y α d y α + y d y y f d f y d y d y y f d f y f d f d d y d y + y d y f d f y d y d y

5 In general, each constrant can be epressed as a functon relatng the varables to each other: g y ; 0 { } and there may be an arbtrary number m of such constrant equatons (well, not qute arbtrary m must be less than, the number of varables n the problem) For any one of the constrant equatons we have: g { y } ; 0 y dg dα η dα y α y Ths last relaton holds even f we multply by an arbtrary functon of : λ ( ) η dα 0 y 0

6 We can sum over all m constrants, and ntegrate from to, and stll get zero: m λ ( ) y ow we can add ths rather eotc-lookng form of the number 0 to our equaton for J : ηdα m J f d f λ ( ) α + + y d y y But snce the varables are related, we stll can t assume that each term n [] s zero ndvdually 0 η d

7 However, we can always redefne the varables of the problem such that the frst -m of them are ndependent The remanng m varables are related by the m constrant equatons For the -m ndependent varables, we must have the terms n [] be zero ndvdually: m f d f + + λ ( ) 0 for < m y d y y Snce we ve sad nothng so far about the value of the functons λ (), we are free to choose ones that are convenent. In partcular, we choose them such that: m f d f + + λ ( ) 0 for > m y d y y

8 Combnng the epressons on the prevous sldes, we arrve at Euler s Equatons when constrants are appled: m f d f + + λ ( ) y d y y 0 for any You may be concerned that there are +m varables n the above equatons, but only equatons avalable But remember that the m equatons of constrant are also avalable So we can solve for everythng ncludng the λ s! Keep n mnd that we made a specal choce for the λ s durng the dervaton. When we apply these deas to physcs we ll see what they represent

9 Eample: Geodesc on a Sphere Let s say we want to fnd the shortest dstance between two ponts, but we have to move on the surface of a sphere between them That means we need to mnmze the ntegral: + + z ( ) ( ) ( ) J ds d + dy + dz y where only paths that satsfy the followng constrant are consdered: d ( ) g, y, z + y + z R 0 f

10 Snce f depends on two varables n ths problem, we have the followng Euler equatons: f d f λ 0 y d y y f d f λ 0 z d yz z The frst equaton gves: d y 0 y 0 d λ + y + z d z 0 λz 0 d + y + z

11 ote that we can solve each of these for λ: d y d z λ y d z d + y + z + y + z y y y + y z z 3/ + y + z ( + y + z ) z z z + y z y 3/ + y + z ( + y + z ) ( ) + + z y y z y y y z z ( ) yz + y + z z z y z z + zy y + zy y zy zy z zy z z yz + yz y + yz z yz z yy z z

12 zy + zy z zy z z yz + yz y yy z z ote that we haven t used any nformaton from the equaton of constrant yet. We can do so f we take a dervatve of g: dg + yy + zz 0 d yy + zz Pluggng ths back n above gves: ( ) ( ) zy + z y zz + yy yz + y z zz + yy zy z y yz y z One can verfy that a plane passng through the center of the sphere ( A + By z ) s the soluton to the above equaton

13 Here s the verfcaton: z A + By; z A + By ; z By [ ] [ ] A + By y A + By y Byy By y Ay + Byy Ay By y Byy By y ote that the tet does the same problem (Eample 6.4) wthout usng λ In ths case that s easer, snce we re not nterestng n what the functon λ s But sometmes for physcsts, λ s qute mportant we ll see why n a few lectures