MA209 Variational Principles

Transcription

1 MA209 Varatonal Prncples June 3, 203 The course covers the bascs of the calculus of varatons, an erves the Euler-Lagrange equatons for mnmsng functonals of the type Iy) = fx, y, y )x. It then gves examples of ths n physcs, namely optcs an mechancs. It furthermore consers constrane moton an the metho of Lagrange multplers. Requre s a basc unerstanng of fferentaton many mensons, together wth a knowlege of how to solve ODEs. Contents Revew of Calculus 2. Functons of One Varable Functons of Several Varables Varatonal Problems 3 3 Dervaton of the Euler Lagrange Equatons 4 3. The one varable - one ervatve case Solutons of some examples Extenson of the Theory More Dervatves Several epenent functons Relatonshp wth Optcs an Fermat s Prncple 0 4. Fermat s Prncple Optcal Analogy Hamlton s Prncple 2 6 Constrants an Lagrange Multplers 3 6. Fnte Dmensons Two mensons n mensons Examples A functonal constrane by a functonal One functonal constrane by a functon Constrane Moton 20 These notes are base on the 20 MA209 Varatonal Prncpals course, taught by J.H.Rawnsley, typeset by Matthew Eggnton. No guarantee s gven that they are accurate or applcable, but hopefully they wll assst your stuy. Please report any errors, factual or typographcal, to m.eggnton@warwck.ac.uk

2 Revew of Calculus. Functons of One Varable Fgure : Graph showng a maxmum at x = a Suppose that x = a s a maxmum of f. Then the graph of f bears some resemblance to that n fgure. Suppose that f s fferentable an that f a) 0. Then we ether have that f a) > 0 or f a) < 0. Conser the former. f fa+h) fa) a) = lm h 0 h an f h > 0 we have that fa + h) fa) > 0 for h small an so fa + h) > fa), but as fa) s a maxmum, f a) > 0 must be mpossble. A smlar argument shows that f a) < 0 s mpossble. Hence our orgnal assumpton s false, an so f a) = 0. However, there are functons f wth f a) = 0 at values of a whch aren t extrema, for example fx) = x 3. We call a pont a where f a) = 0 a crtcal pont of the functon f an we have shown that the set of extrema s a subset of the set of crtcal ponts. Ths s also true for the set of local extrema. Example. Let fx) = ax 2 + bx + c wth a 0. Then f x) = 2ax + b wth b ) 2a = ax 2 + bx + c a ) b 2 2a b b 2a crtcal pont. fx) f b 2a a x + b 2a) 2 an so for a > 0 s a mnmum an a < 0 a maxmum. the only ) c = ax 2 + bx b2 4a + b2 2a = In general, ths won t be so pretty, but for nce functons wth Taylor seres we have fa + h) fa) = hf a) + h2 2! f a) +... an so f f a) 0 we can ece f f a) > 0 whence we have a local mnmum an f f a) < 0 we have a local maxmum..2 Functons of Several Varables We wll look at the two varable case. Conser f, x 2 ) a fferentable functon wth extremum a, a 2 ). Pck functons t) an x 2 t) such that 0) = a an x 2 0) = a 2. Set gt) = t), x 2 t)). Then g takes some values of f an at t = 0 s an extremum of f an hence g. Thence g 0) = 0 an so f a, a 2 ) s an extremum then we have that t ft), x 2 t)) = 0 ) t=0 for any par of functons t), x 2 t)) passng through a, a 2 ). Thus by the chan rule we have that a, a 2 ) 0) + a, a 2 ) x 2 t x 2 t 0) = 0 As ths s true for arbtrary functons, we must have that a, a 2 ) = 0 = x 2 a, a 2 ). Note that we coul have pcke functons wth nepenent ervatves at t = 0 specfcally. 2 of 20

3 For n varables, fx) real value an wth an extremum at x = a, we pck a functon g v t) = fa + tv), where v s an arbtrary vector. Then ths wll have an extremum at t = 0 so g v0) = 0 for all v an so fa) v = 0 for all v so fa) = 0. If t = 0 s a local maxmum of g v then a s a local maxmum of f. Then g v0) = j v v j x x j a) = Hess f a) < 0 Then all egenvalues of Hess f a) must be negatve. If they have mxe sgns or are zero then we can euce nothng. Example.2 Suppose that fx, y) = ax 2 +bxy+cy 2. Then f = 2ax+by, bx+2cy) = 0, 0) for an extrema. Thus f 4ac b 2 0 then x = 0 = y s the only crtcal pont. 2 f 2 f v v j a) = v x x 2a 2 + 2bv v 2 + 2cv2 2 j j an so f a 0 we get ths equal to 2 a v + b 2a v ) 2 ) ) 2 + c b2 4a v2 2 ) maxmum or mnmum when a c b2 4a > 0 an 4ac > b 2. 2 Varatonal Problems an so we have a In orer to motvate the stuy of Varatonal Prncples we gve some examples of famous problems n the subject.. Suppose that y s a functon such that y ) = y an yx 2 ) = y 2. We want to fn y wth the shortest length. The length Ly) s gven by ) y 2 Ly) = + x x We say that L s a functonal of the functon y 2. Brachstochrone. Suppose that we have a bea of mass m slng own a frctonless wre uner gravty along a curve from, y ) to x 2, y 2 ). Let T y) be the tme taken to go from, y ) to x 2, y 2 ) along the curve y. We want to fn a mnmum of ths. If the tme s t at, y ) an t 2 at x 2, y 2 ), an we enote by s the arclength parametrsaton, then ) 2 s2 s2 s x2 + y T y) = t 2 t = t = = t s s v = x x v s s t We can fn the velocty v from conservaton of energy. We know that E = 2 mv2 + mgyx) y ) = 2 mv2 + 0 f the ntal spee s v. If we set v = 0 then v = 2gy yx)) an so T y) = ) 2 y + x 2gy yx)) x 3. Least area of revoluton Take a curve y wth y ) = y an yx 2 ) = y 2 an rotate t about the x-axs. One then gets a surface of revoluton aroun the x-axs. We want to fn the curve for whch the surface area s as small as possble. The surface area s equal to Ay) = 2π y + y ) 2 x 3 of 20

4 3 Dervaton of the Euler Lagrange Equatons 3. The one varable - one ervatve case The problems n secton 2 nvolve mnmsng functonals bult from a functon of one varable by ntegraton of the functon an ts ervatves wth values of the functon specfe at the ens of the range of ntegraton. These are typcally calle fxe enpont problems. In general, the class of problems of ths kn have a functonal of the form Iy) = fx, yx), y x))x 2) for yx) wth y ) = y an yx 2 ) = y 2. In future I wll wrte y for yx) an y for y x) to smplfy the notaton. How o we fn extrema of Iy)? We procee n a smlar manner to fnng contons for functons at extrema. We conser a one parameter famly of functons y t wth y 0 the extremsng functon. Clearly they all have the same fxe enponts. Then f gt) = Iy t ) we have g 0) = 0 or t Iy t) t=0 = 0. If y t = y 0 + tv then v ) = 0 = vx 2 ) Hence t Iy 0 + tv) t=0 = 0 3) for v as efne above. The solutons to ths equaton are calle crtcal ponts of Iy). Example 3. Conser We then have an so t Iy 0 + tv) = t=0 Iy 0 + tv) = 0 Iy) = 0 0 [xy 2 + y ) 2 ]x [xy 0 + tv) 2 + y 0 + tv ) 2 ]x [xy 0 + tv) 2 + y 0 + tv ) 2 ]x = t=0 0 2xy 0 v + 2y v )x y 0 s a crtcal pont f the ntegral s 0 for all v wth the contons as above. In the general case Iy 0 +tv) = x 2 fx, y 0 + tv, y 0 + tv )x an so f we procee formally we get t Iy 0 + tv) = fx, y 0 + tv, y 0 + tv )x t=0 t x t=0 x2 = y x, y 0, y 0)v + y x, y 0, y 0)v x If y 0 s a crtcal pont an vx) s any sutable functon wth v ) = 0 = vx 2 ) then we have from equaton 3) 0 = = = x x2 y x, y 0, y 0)v + y x, y 0, y 0)v x [ y )] x y vx + y v x 2 x [ y )] x y vx 4 of 20

5 an hence we want to solve [ y )] x y vx = 0 for sutable v. We now make rgorous sense of ths, an so we nee ) f an ts partal ervatves up to orer two an y 0 to be contnuous. Then y t y s contnuous. We also nee y 0 + tv to be a famly of functons n a sutable space an so v must have two contnuous ervatves. Theorem 3. The Funamental Theorem of the Calculus of Varatons) If ux) s contnuous on [, x 2 ] an ux)vx)x = 0 for all vx) wth two contnuous ervatves an v ) = 0 = vx 2 ) then ux) = 0 for all x [, x 2 ]. Proof We use a contracton argument. Suppose there s some pont x 0, x 2 ) wth ux 0 ) 0. Wthout loss of generalty we can assume that ux 0 ) > 0. IF not, conser the functon u. Then ux) s non zero on some nterval aroun x 0 postve here even), as u s contnuous. Call ths nterval x, x 2 ). Suppose we have vx) wth two contnuous ervatves an vx) = 0 where x [x, x 2 ]. Then 0 = ux)vx)x = x 2 x ux)vx)x If furthermore vx) > 0 for any x x, x 2 ) then we have that x 2 x ux)vx)x > 0 Ths s a contracton; hence there s no pont x 0 where ux 0 ) 0 an so ux) = 0 for all x [, x 2 ]. Thus the proof s reuce to a constructon of such a functon vx). A sutable functon woul be { 0 x x vx) =, x 2 ) x x )3 x x 2 )3 x x, x 2 ) Q.E.D. Remark If functonals have more ervatves then ths argument coul be mofe for those. We smply take one hgher power than the ervatves. Ase If we nee nfntely many ervatves, we can use e x 2 as t has nfntely many ervatves at x = 0 an they are all equal to zero. Theorem 3.2 If f s a functon of three varables wth all partal ervatves up to orer two contnuous then any crtcal pont y of Iy) = x 2 fx, yx), y x))x on the set of functons wth two contnuous ervatves an satsfyng enpont contons y ) = y an yx 2 ) = y 2 has y ) x y = 0 x [, x 2 ] 4) 5 of 20

6 Proof We showe above that [ y )] x y vx = 0 for all v wth two contnuous ervatves. The expresson n the square brackets s contnuous an so by the funamental theorem theorem 3.) must be zero x [, x 2 ] Q.E.D. Defnton 3. If a functonal Iy) = x 2 fx, yx), y x))x then f s calle the Lagrangan of I an y x y ) = 0 s calle the Euler-Lagrange equaton of I Remark The E-L equaton s a secon orer ODE for yx) wth enpont contons. 3.2 Solutons of some examples Example 3.2 Fn the E-L equaton for Iy) = π 2 0 y2 y ) 2 )x. We have that y = y an y = y an so the E-L equaton s y x y ) = 0 gvng y + y = 0 We now solve the examples n secton 2.. We have from before that Ly) = + ) y 2 x x an so y an so = 0 an y = y +y ) 2 y. The E-L equaton then gves +y ) 2 ) y = 0 + y ) 2 s constant, hence y = m gvng the lne y = mx + a Remark Any case where x ) y y = 0 wll have an mmeate ntegral of the E-L equaton = 0 as y = constant. We call ths a frst ntegral of the E-L equaton. Before lookng at the other two examples, we note that x oes not appear explctly so we ask f there s a frst ntegral. Observe that y ) x y f = y y + y x y = y x y ) y x y an f y s a soluton of the E-L equatons we have that y ) x y f = x ) + y + y x y y an so f f s nepenent of x then y y f s a constant. Ths s calle the frst ntegral for the case of a Lagrangan nepenent of x. 6 of 20

7 2. Brachstochrone We have fx, y, y ) = +y ) 2 y y) no x epenence here an so y y f s a constant. gvng f we gnore the constants. There s y y f = y y + y ) 2 y y) + y ) 2 y y) = A +y ) 2 = A an hence + y y) y ) 2 )y y) = A 2 y = ± A 2 y y) an we thus get If we now make the substtuton A 2 y y) = sn 2 θ 2 then we get that A2 y = sn θ 2 cos θ 2 θ an we get that A 2 sn θ 2 cos θ sn 2 θ 2 θ 2 = ± sn 2 θ 2 = ± cos θ 2 sn θ 2 an so A 2 sn 2 θ 2 θ = ± gvng 2A 2 cos θ)θ = ± an then ntegratng gves θ sn θ) = B ± x 2A2 whch mplctly etermnes θx) an so yx) Ths curve s calle a cyclo. Fgure 2 shows such a curve. Fgure 2: A cyclo 3. fx, y, y ) = 2πy + y ) 2 an observe that we have no x epenence agan. Thus we look at the frst ntegral: y y f = y ) 2 y + y ) 2 y + y ) 2 = A an so we get that y + y ) 2 = A an so y y = ± 2. If we then make the substtuton y A 2 A y A = sn zz an hence the equaton to solve becomes A snh zz = ± cosh 2 z = ± snh z = cosh z we get that 7 of 20

8 Fgure 3: The shape of surface whch mnmses the surface of revoluton an thus we get that z = ± A an so z = B ± x A an so y = A cosh B ± x ) A an so t looks lke fgure 3 We now try to ft ths shape of soluton to the enpont contons. Wthout loss of generalty we wll assume that y = A cosh B + A) x, an we want a soluton wth y ) = y an yx 2 ) = y 2. Usng the frst of these we get that B = cosh y ) A A an then y = y cosh ) x A + y 2 A 2 snh ) x A an usng the secon conton gves a pretty nasty equaton I leave to the reaer to work t out). To see f solutons exst we plot the graph of yx) for varous values of A. Thus from ths graph you can see that f x 2, y 2 ) s to the rght of the otte lne then there s no soluton. Also note that f x 2, y 2 ) s above the otte lne then there are two solutons. Also these solutons may not be extrema, as a broken lne may well mnmse the problem. Remark y 0 + tv s calle a varaton of y 0, hence the name Calculus of Varatons 3.3 Extenson of the Theory 3.3. More Dervatves Suppose that Iy) = fx, y, y,..., y n) )x We try the same metho as before, conserng Iy + tv) for y an extremum. Set gt) = T y + tv) an then ths has an extremum at t = 0 so g 0) = 0 an thus t Iy + tv) t=0 = 0 an so x2 [ Iy + tv) t = t=0 y v + y v ] vn) x = 0 y n) If we assume that v ) = 0 = vx 2 ) an all partal ervatves up to v n ) are zero at an x 2 then we get that x2 [ t Iy + tv) t=0 = y ) )] x y ) n n x n y n) vx = 0 For the argument to be complete we nee f to have n + ) contnuous ervatves an y to have 2n contnuous ervatves. Then the term n square brackets s contnuous an we nee 8 of 20

9 the verson of the funamental theorem for v wth 2n contnuous ervatves. Then for y an extremum t satsfes y ) ) x y ) n n x n y n) = 0 5) Ths s agan calle the Euler Lagrange equaton for the functonal. There s no exstence or unqueness theorem n ths case agan. Example 3.3 Suppose Iy) = π 2 0 y ) 2 y 2 )x wth y0) = 0 = y 0) an y π 2 ) = an y π 2 ) = 0. 2 The E-L equaton gves 2y + 2y ) = 0 an so y 4) y = 0 an ths x 2 has a general soluton of y = A cos x + B sn x + Ce x + De x an solvng for the enpont contons gves the four equatons 0 = A + C + D, 0 = B + C D, = B + Ce π 2 + De π 2 an 0 = A + Ce π 2 De π 2 an these can be solve Several epenent functons Problems nvolvng curves may not be expressble as y = yx) an so nstea we coul wrte the curve n parametrc form,.e. for the length problem we coul wrte Lx, y) = x ) 2 + y ) 2 t. In general these have the form t Ix, y) = t ft, xt), yt), x t), y t))t an we use a one parameter varaton x + hu, y + hv). Then x, y) s an extremum of I means that Ix + hu, y + hv) h = 0 h=0 Note that u an v must vansh at the enponts to preserve the enpont contons. If we frst take vx) = 0 x [t, t 2 ] then h Ix + hu, y) h=0 = 0 an so x ) t x = 0 Smlarly f ux) = 0 x [t, t 2 ] then h Ix, y + hv) h=0 = 0 an so y ) t y = 0 In other wors both x an y satsfy the Euler Lagrange equaton for one varable. One can also erve these two equatons as we before: usng the chan rule on the necessary conton, then ntegratng by parts. Then takng v = 0 an then u = 0 we can apply the funamental theorem n both cases, gvng the result above. It shoul be clear that ths works for any number of nepenent varables, so long as they can be vare nepenently. If I,..., x n ) = then I has n smultaneous E-L equatons t ft, t),..., x n t), x t),..., x nt))t ) = 0 =,..., n 6) x t x 9 of 20

10 Example 3.4 Suppose that Lx, y) = 0 x ) 2 + y ) 2 t wth x0) = an x) = x 2 as well as y0) = y an y) = y 2. Ths has two E-L equatons: ) x t = 0 x ) 2 +y ) 2 ) y = 0 x ) 2 +y ) 2 t x an so both an y are constants. Thus x ) 2 +y ) 2 x ) 2 +y ) 2 x ) 2 +y ) x, y ) = A, B) 2 s a constant unt vector. Hence xt), yt)) s a curve wth a constant recton. If ct) = x ) 2 + y ) 2 then x, y) = t)a, B) + C, D) where = c Remark Observe that although t s wrtten n term of two varables, the problem s egenerate. It has nfntely many solutons gven by fferent possble functons t). If there s no explct t epenence,.e. t = 0, then conser F t) = x x x n x n f Then F t = x = 0 x ) + x t x +... ) + x n x + x n n t x t x... x n x... x n x n x x n f there s no explct tme epenence an,..., x n satsfy the E-L equatons. Hence F s constant an ths s another Frst Integral. 4 Relatonshp wth Optcs an Fermat s Prncple We look here at rays of lght n the plane movng wth spee cx, y) 4. Fermat s Prncple Theorem 4. Fermat s Prncple) Lght Travels along a path between two ponts, y ) an x 2, y 2 ) so as to take the least tme to get from, y ) to x 2, y 2 ) cx, y) s the spee at x, y), an f we travel along a path the spee wll be the rate of change of arclength along the path. Thus f we measure arclength s from an ntal poston, then s t = cx, y). If the path s a graph of a functon yx) then from a path from, y ) to x 2, y 2 ), where we are at, y ) at tme t an arclength s an at x 2, y 2 ) at tme t 2 an arclength s 2, we get that T y) = t 2 t = t t = s2 s s s t = s2 s s x2 + y c = ) 2 x cx, y) The actual path followe by a lght ray wll be a mnmum of T y). 0 of 20

11 Example 4. Lght n a homogeneous meum Here we assume that c s a constant. We have that T y) = + y c ) 2 x = c Ly) an hence n a homogeneous meum lght travels n straght lnes snce these are crtcal ponts of the length functonal. Example 4.2 The Law of Refracton Suppose that we have two homogeneous mea wth spees c an c 2 an have a straght lne nterface an a ray of lght from the frst to the secon. We know that we wll have a broken lne, but what s the change n recton at the nterface. We look at broken straght lne paths passng through the pontx 0, 0) on the x-axs. The tme taken s τx 0 ) an ths s equal to τx 0 ) = x0 ) 2 + y 2 c + x2 x 0 ) 2 + y 2 2 The actual path wll be a mnmum wth respect to x 0 an so at that pont where the path crosses the x-axs we have τ x 0 = 0. Now τ x x 0 = 0 x 2 x 0 = 0 whence sn θ c sn θ 2 c 2 = 0 or Ths s known as Snell s Law. c 2 c x0 ) 2 +y 2 c 2 x2 x 0 ) 2 +y 2 2 sn θ sn θ 2 = c c 2 7) Suppose that c s only a functon of y,.e. that cx, y) = cy). We ve nto strps parallel to the x-axs. In each strp, the path s approxmate by a straght lne segment. Then the slope n the strp wll be approxmately y y x. We then have that cot θ = x = y an then sn θ =. It s cot θ here because θ s the angle the ray makes wth the y 2 +y ) recton. Accorng to Snell s Law sn θ c s a constant an so cy) +y ) 2 s a constant. Ths equaton gves + y ) 2 = Kcy) an so y = ±. Then vng by the square K 2 c 2 y) root term an ntegratng wth respect to x gves y = A ± x K 2 c 2 y) Ths gves an equaton for x as a functon of y an by solvng, or usng a substtuton, we get an explct soluton. We now rework the above usng the Calculus of Varatons. In ths case we have a functonal nepenent of x as T y) = + y ) 2 x cy) an then ths has a frst ntegral of y y cy) + y ) 2 + y ) 2 cy) = K an ths gves cy) +y ) 2 ntegral of Fermat s Prncple s Snell s Law. = K whch we euce from Snell s law before. Hence the frst of 20

12 4.2 Optcal Analogy If a problem n the Calculus of Varatons leas to a functonal of the same form as that comng from Fermat s Prncple an the optcal problem s alreay solve then the same soluton apples to the varatonal problem. It then has the soluton, when nepenent of x n the functonal, gven by y A ± x = K 2 c 2 y) Ths was how Bernoull frst solve the Brachstochrone problem, where we have + y ) 2 2gy y) x as our functonal. If we take cy) = 2gy y) then we can wrte own the ntegral formula for the soluton. When x appears explctly we have to go to the full E-L equatons. 5 Hamlton s Prncple Suppose that xt) = xt), yt), zt)) escrbes the moton of a pont partcle n three mensons where t s the tme varable. We efne ẋ := x t an call t the velocty v. Furthermore we efne ẍ := 2 x an call t the acceleraton. v := v := ẋ t 2 + ẏ 2 + ż 2 = v v 2 s calle the spee. The moton s governe by the mass m > 0. The knetc energy s 2 mv2 = 2 mẋ2 + ẏ 2 + ż 2 ). If we have many partcles then the total knetc energy s T = 2 m v 2. If q,..., q n s a fferent set of coornates of whch, y, z, x 2, y 2, z 2,... are functons then we get T as a functon of q,..., q n, q,..., q n by substtuton. Defnton 5. A conservatve system s where the forces actng F can be gven n terms of a functon V such that F = V. V s calle the potental energy an s a functon of q,..., q n nepenent coornates. Defnton 5.2 The Lagrangan of the system s Lq,..., q n, q,..., q n ) := T V Example 5. Suppose that a partcle of mass m s movng n a crcle n the x-y plane wth gravty actng n the negatve y recton. Then the potental s gven by V := mgy = mgr sn θ an the knetc energy s T = 2 mr2 θ2 an so the Lagrangan s Lθ, θ) = 2 mr2 θ2 mgr sn θ Theorem 5. Hamlton s Prncple) The path followe by a system escrbe by a Lagrangan L = T V n gettng from an ntal poston P at tme t to a fnal poston P 2 at tme t 2 s a crtcal pont of the functonal I = t Lt amongst all possble paths from P to P 2 at the relevant tmes. Hence the actual path satsfes the E-L equatons for L, namely L ) L = 0 for =,..., n 8) q t q 2 of 20

13 Example 5.2 Suppose we have a partcle on a crcle of raus R an s acte upon by gravty see example 5.). Then we have Lθ, θ) = 2 mr2 θ2 mgr sn θ an so the E-L equatons for ths gves mgr cos θ t mr2 θ) = 0 = θ + g R cos θ = 0 an ths s calle the penulum equaton Example 5.3 Suppose that we have a partcle of mass m movng n R 3 wth a force F = V. Then L = 2 mẋ2 + ẏ 2 + ż 2 ) V x, y, z) an the E-L equatons gve L x L t ẋ L y L t ẏ L z L t ż ) = 0 = V ) x mẍ = 0 = 0 = V y ) mÿ = 0 = 0 = V z m z = 0 = F mẍ = 0.e. Newton s Secon Law. Thus Hamlton s prncple s n accor wth Newton s secon Law. Observe that L s nepenent of the tme varable, an so we always have a frst ntegral of the form L L q q n L = constant q q n Observe that the knetc energy s quaratc n the ervatves, an wll be so for any system. Thus n n T q,..., q n, q,..., q n ) = q q j T j q,..., q n ) An hence we get the entty = j= T q,..., q n, aq,..., aq n ) = a 2 T q,..., q n, q,..., q n ) 9) whch s calle Euler s Formula. It shoul be clear that L q q,..., q n, the frst ntegral becomes = T q as V s nepenent of the T T q q n L = constant q q n an hence, by fferentatng 9) wth respect to a an settng a = 0, we get that T + V = constant an ths s calle conservaton of energy. 6 Constrants an Lagrange Multplers 6. Fnte Dmensons 6.. Two mensons A typcal example s to fn extrema of fx, y) on the set {x, y) R 2 gx, y) = 0}. The mplct functon theorem tells us whch varable n an equaton can be solve for n terms of the others. If g x x 0, y 0 ) 0 for some pont then there s a functon ηx) efne for x near x 0 wth ηx 0 ) = y 0, η fferentable, such that all solutons x, y) of gx, y) = 0 near x 0, y 0 ) have the form x, ηx)). We say the constrant s regular f at every soluton at least one of the partal ervatves s non-zero. 3 of 20

14 If x 0, y 0 ) s an extremum of f on {x, y) gx, y) = 0} then let y = ηx) be a soluton near x 0, y 0 ) of the constrant, an then substtute ths n f to gve fx, ηx)) an ths has x 0 as an extremum. Therefore fx, ηx)) x = 0 0) x=x0 We also have the fact that gx, ηx)) = 0 for all x for whch η s efne. Equaton 0), by the chan rule, yels x x 0, y 0 ) + y x 0, y 0 ) η x x 0) = 0 an we also have that x gx, ηx)) = 0 = g x g η x, y) + x, y) y x x) = 0 for all x near x 0, an f one evaluates ths at x = x 0 we get that g η x x x 0) = x 0, y 0 ) g y x 0, y 0 ) as the enomnator s non zero by assumpton. From these we get that: an f we efne λ = y x 0,y 0 ) g y x 0,y 0 ) x x 0, y 0 ) y x 0, y 0 ) then ths becomes x x 0, y 0 ) λ g x x 0, y 0 ) = 0 = g x x 0, y 0 ) g y x 0, y 0 ) = 0 x f λg) x0,y 0 ) = 0 Also, by efnton of λ we get that y f λg) x0,y 0 ) = 0 an therefore f λg has a crtcal pont at x 0, y 0 ). Smlarly f g y x 0, y 0 ) 0 an x 0, y 0 ) s an extremum of f on {x, y) gx, y) = 0} then there s a constant λ = x x 0,y 0 ) g x x 0,y 0 ) such that f λ g has a crtcal pont at x 0, y 0 ). We have thus prove: Theorem 6. Lagrange Multpler) If g s a regular constrant wth g 0 for all x, y) wth gx, y) = 0 then any extremum x 0, y 0 ) of fx, y) on the set {x, y) gx, y) = 0} has an assocate real number λ such that f λg has a crtcal pont at x 0, y 0 ). We call λ the Lagrange multpler for x 0, y 0 ). The unknowns are now x 0, y 0 ) an λ. The conton of f λg havng a crtcal pont at x 0, y 0 ) s f λg)x 0, y 0 ) = 0 an we also have the conton of gx 0, y 0 ) = 0. Example 6. Fn the extrema of fx, y) = ax + by on x 2 + y 2 =. The extrema has a crtcal pont of f λg = ax + by λx 2 + y 2 ) an so 0 = a 2λx an 0 = b 2λy gvng a 2 + b 2 = 4λ 2 an so λ = ± a 2 +b 2 2 an so x, y) = ± a, ± b a 2 +b 2 a ). Then 2 +b 2 f = ± a2 +b 2 = ± a a 2 + b 2 an hence there s a maxmum at + a 2 + b 2 an a mnmum at 2 +b 2 a 2 + b 2. In general there may be more solutons to f λg)x 0, y 0 ) = 0 an gx 0, y 0 ) = 0 than there are extrema x 0, y 0 ) of f on {x, y) gx, y) = 0}. Defnton 6. We call the solutons to the above constrane crtcal ponts of f The constrane crtcal ponts of f on gx, y) = 0 are unconstrane crtcal ponts of f λg for some λ. 4 of 20

15 6..2 n mensons Let f be a functon of n varables an look for extrema of f,..., x n ) on the set of ponts where a functon g,..., x n ) = 0. Suppose that x =,..., x n ) s an extreme pont an pck two vectors u an v an conser a functon of two varables F u,v h, k) := fx+hu+kv) subject to the constrants G u,v h, k) := gx + hu + kv) = 0. Then h, k) = 0, 0) s an extremum of F u,v subject to G u,v h, k) = 0. Hence there s a Lagrange multpler λ u,v such that F u,v λ u,v G u,v has a crtcal pont at h, k) = 0, 0). Therefore we have that an that Ths then gves us that an that h F u,v λ u,v G u,v ) = 0 k F u,v λ u,v G u,v ) = 0 u f λ u,v g)x) = 0 v f λ u,v g)x) = 0 from the fact that Fu,v h = h fx + hu + kv)) = u fx + hu + kv) an smlarly for the other partal ervatves. Then for every par of vectors u an v we have that f λ u,v g)x) s perpencular to both u an v. If e,..., e n s the stanar bass an λ j = λ e,e j then fx) λ j gx) s perpencular to both e an e j for each an j. In terms of the partal ervatves ths becomes x x) λ j g x x) = 0 = x j x) λ j g x j x), j We am to fn a conton that s nepenent of j an so have a sngle Lagrange multpler for each of the equatons. For a regular constrant we nee g 0 everywhere on gx) = 0. Thus at least one partal ervatve of g s non zero, say the 0 th. Then we can wrte λ 0 j = x 0 x) g x 0 x) an ths s nepenent of j. Then f we put λ := λ 0 j for any j an then nput ths nto the secon equaton, we get that x) λ g x) = 0 j x j x j Hence we have a λ such that f λg)x) = 0 Example 6.2 Fn the pont on the plane x n = p closest to a gven pont a not on the plane. We am to mnmse the stance from a pont x to the pont a such that x n = p The Euclean stance s gven by x, a) = x a but we wll take the square of ths to smplfy workng out. It shoul be clear that f the square of the stance has a mnmum, then so must the stance tself. Thus we have that we want to mnmse fx) = x a 2 = =m = x a ) 2 subject to gx) = x n p. Now f = 2x a) an g = n. At the crtcal pont there s a number λ such that f λg = 0 an n ths case ths s 2x a) λn = 0 an so we get that x = a + λ 2 n an a + λ 2 n) n = p an so λ 2 = p a n thus x = a + p a n)n. Ths s a mnmum as any pont whch s fferent from the above one wll have stance on a hypotenuse of a rght angle trangle wth one se equal to the length at a crtcal pont. 5 of 20

16 6..3 Examples The followng examples are ones that we am to solve, an wll evelop technques to o so n the next secton.. Hangng rope or chan Suppose we have a rope hangng n equlbrum between two ponts, y ) an x 2, y 2 ). What s the shape of the rope? Ths s calle a catenary. Suppose the shape s the graph of a functon y = yx). In equlbrum ts potental energy wll be mnmse. Let ρ be the ensty per unt length of the rope an assume that t s constant. Then the total mass s Jy) := ρ x 2 + y ) 2 x =: M. The potental energy s then gven by Iy) := ρg x 2 y + y ) 2 x. Hence we want to mnmse Iy) subject to Jy) beng a constant value M. 2. Isopermetrc problem Conser a close curve n the plane. For a gven length, we want to fn the curve whch encloses the greatest area. Let the curve C be gven by xt), yt)) wth xt ) = x 0 = xt 2 ) an yt ) = y 0 = yt 2 ). Then the length of C s gven by LC) = t ẋ2 2 + ẏ 2 t an the area s gven by AC) = 2 t xẏ yẋ)t We want to mnmse AC) for fxe LC). 3. Geoescs on Surfaces Curves whch mnmse the stance n a surface are calle geoescs. Here we mnmse a length functonal Lx) for curves xt) whch satsfy gxt)) = 0 for all t A functonal constrane by a functonal We frst look at problems wth two functonals lke one an two above), wth two parameter varatons, an then look at two varaton problems. If Iy) s extremse on the set gx) wth Jy) constant, we look at two parameter varatons y +hu+kv where u an v are chosen such that u ) = ux 2 ) = v ) = vx 2 ) = 0 an then h, k) = 0, 0) s an extremum for Iy + hu + kv) = J 0 whch s a fxe constant. Defne F uv h, k) = Iy + hu + kv) an G uv h, k) = Iy + hu + kv) J 0. Then F uv has an extremum at h, k) = 0, 0) for h, k) such that G uv h, k) = 0. Hence we have a Lagrange Multpler λ uv such that F uv λ uv G uv has a crtcal pont at 0, 0). Thus an also h F uvh, k) λ uv G uv h, k)) h,k=0 = 0 k F uvh, k) λ uv G uv h, k)) h,k=0 = 0 If Iy) = x 2 fx, y, y )x an Jy) = x 2 gx, y, y )x then the h partal equaton gves 0 = y f λ uvg) )) x y f λ uvg) ux an the k partal equaton gves Then we have that 0 = 0 = y f λ uvg) x t )) y f λ uvg) vx y )) x2 g x y ux λ uv y )) g x y ux 6 of 20

17 The regularty conton gves that the latter ntegran an hence ntegral n the above equaton s non zero on the set of gx) an so Jy) = J 0. Then the former ntegral s non zero an so we can set )) x2 y x y u λ u0 v = 0 x )) g y g x y u 0 x an note that the rght han se here s nepenent of v, an so we can wrte λ u0 v =: λ. Then for any v vanshng at an x 2, an for λ efne before, we get that )) f λg) f λg) vx = 0 y x y an by the funamental lemma we get that f λg) y x ) f λg) = 0 y Ths s calle the Euler Lagrange equaton for ths case. We have thus prove: Theorem 6.2 An extremum of Iy) subject to Jy) = J 0 satsfes the Euler Lagrange equaton ) f λg) f λg) = 0 ) y x y for I λj for some λ calle the Lagrange Multpler. Remark Ths proof can be aapte to more ervatves or more nepenent varables. We now solve the examples gven at the start of ths subsecton.. Catenary We have Iy) := ρg x 2 y + y ) 2 x an Jy) := ρ x 2 + y ) 2 x =: M. y satsfes the E-L equaton for I λj for some λ. Ths functonal s ρ gy λ) + y ) 2 x an we use the optcal analogy to solve t. Ths correspons to lght movng wth spee c = gy λ an has soluton of y = λ + c ρgx cosh + c 2 ρg c an we have three contons an three unknowns an so we can solve to fn c, c 2, λ 2. Isopermetrc Problem We want to maxmse Ax, y) whle keepng Lx, y) = l fxe. xt), yt)) s a parametersaton of a close curve. The extremsng curve wll satsfy the E-L equatons for A λl an so we get ) A λl)x, y) = 2 xẏ yẋ) λ ẋ 2 + ẏ 2 t an we have two E-L equatons an so we have ) 2ẏ t 2 y λẋ2 ẋ = 0 +ẏ 2 ) 2ẋ t 2 x λẋ2 ẏ = 0 +ẏ 2 t 7 of 20

18 Note that both equatons are tme ervatves an so we get ) ẋ t y λ ẋ2 = 0 +ẏ 2 ) ẏ x + λ ẋ2 = 0 +ẏ 2 t an ntegratng once gves y λ x + λ ẋ2 ẋ +ẏ 2 ẋ2 ẏ +ẏ 2 = B = C an hence we get that x C) 2 + y B) 2 = λ 2 an ths s a crcle centre C, B) an raus λ. Therefore 2πλ = l an so λ = l 2π. To hanle example three we nee a new metho One functonal constrane by a functon As far as I can gather, we cannot n general constran a functonal wth respect to a gven functon, but we can o f the functonal s a functon of curves. Suppose we have curves xt) = t),..., x n t)) jonng two ponts x ) an x 2) at tmes t an t 2 an the curves satsfy gt, xt), ẋt)) = 0. We have some functonal Ix) = t2 t ft, xt), ẋt))t an we am to extremse amongst these curves. An example of ths s fnng geoescs n a surface. Let x h be a varaton of an extremum x, an x h = x + hu + oh 2 ) whch satsfy the constrant for all h. Then h = 0 s a crtcal pont for Ix h ) as a functon of h an so h Ix h) = 0 h=0 an thus we get that t [ n u + ) ] u t = 0 2) x x = Dfferentatng the constrant gt, xt), ẋt)) = 0 gves n g u + g ) u = 0 for all t 3) x x = Pck a functon λt), multply 3) by λ an subtract from the ntegran of 2). Therefore we get 0 = = t = t n u + x [ n = x f λg)u + g u λt) x u + g )) u t x x ) ] f λg) u t x Observe that ut ) = 0 = ut 2 ) an then ntegratng the u terms by parts gves [ t2 n 0 = f λg) )) ] f λg) u t x t x t = 8 of 20

19 We can pck λt) so that one of the) coeffcents, say of u, s zero. We can o ths snce settng f λg) t x f λg) = 0 gves a frst orer lnear nhomogeneous ODE for λ whch can be solve by the ntegratng factor metho. The constrant 2) amongst the u then etermnes u n terms of u 2,..., u n an then the latter can be vare freely. Hence the conton for an extremum becomes n [ 0 = f λg) )) ] ) f λg) u t x t x =2 t an snce now u 2,..., u n are arbtrary, vanshng at t an t 2 we can apply the funamental theorem to each u n turn takng the rest of u 2,..., u n to be zero gvng f λg) ) f λg) = 0 for = 2,..., n x t x For = ths equaton was the way we chose λt) an hence the equaton becomes We have thus prove f λg) ) f λg) = 0 for =, 2,..., n 4) x t x Theorem 6.3 To extremse a functonal I gven by a Lagrangan f amongst curves xt) wth fxe enponts subject to a constrant gt, xt), ẋt)) = 0 there s a functon λt) such that the Euler-Lagrange equatons for the Lagrangan f λg namely 4)) are satsfe. Example 6.3 Geoescs on a surface n R 3 gven by an equaton gx) = 0. Geoescs are paths of shortest length an so mnmse t ẋ2 2 t + ẏ 2 + ż 2 t. Then there s a functon λt) such that a geoesc xt) satsfes the E-L equatons of the Lagrangan ẋ 2 + ẏ 2 + ż 2 λt)gx). We thus have, for the x equaton, ) λt) g x ẋ t ẋ2 = 0 + ẏ 2 + ż 2 We am to smplfy ths equaton, an so we ntrouce the arclength parameter s. Ths s gven by s t = ẋ 2 + ẏ 2 + ż 2 an we change nepenent varables from t to s. Then ẋ s =. Then vng the equaton by 2 t an puttng µ = λ s we get that x s = ẋs t ẋ2 +ẏ 2 +ż an the y equaton becomes an smlarly the z equaton becomes µ g x 2 x s 2 = 0 µ g y 2 y s 2 = 0 t an so µ = µ g z 2 z s 2 = 0 2 x s 2 g x = 2 y s 2 g y = 2 z s 2 g z 9 of 20

20 There s no general metho to solve these equatons. We now conser a specal case of a sphere n R 3, an so we have that gx, y, z) = x 2 + y 2 + z 2 R 2. Ths then gves us that an notce that s z y s y z s ) 2 x 2x x 2 = 2 y 2y x 2 = 2 z 2z x 2 = z y s s + z 2 y s 2 y z s s y 2 z s 2 = yz 2 y y s 2 2 ) z z s 2 = 0 an so z y s y z z s = A s constant. Smlarly x s z x x s = B s constant an y s x y s = C s constant. Therefore multplyng by x, y an z respectvely on these equatons gves 0 = Ax + By + Cz. Ths s a plane through the orgn perpencular to A, B, C). Hence the path must le n the ntersecton of the sphere an a plane through the orgn. These are calle Great Crcles. We have two solutons to the E-L equatons satsfyng the enpont contons so long as the enponts are not antpoal. If two enponts are poles then we get a contnuum of great crcles all of whch are solutons to the problem. 7 Constrane Moton For partcles movng wth coornates relate by a constrant, say g = 0, then Hamlton s prncple extremses t 2 t Lt where L = T V an now we are subjecte to a constrant. We use the Lagrange Multpler metho, an so we have a functon λt) such that the moton satsfes the E-L equaton for T V λg Example 7. Conser free moton on a surface n R 3. We then have by efnton V = 0 an also T = 2 mẋ2 + ẏ 2 + ż 2 ) an gx, y, z) = 0. Thus we want the E-L equatons for 2 mẋ2 + ẏ 2 + ż 2 ) λt)gx, y, z) an so we have that λ g x t mẋ) = 0 λ g y t mẏ) = 0 = mẍ = λ g λ g z t mż) = 0 g s a vector perpencular to the surface at each pont. If we elmnate λ m then we get that Observe that ẍ g x = ÿ g y = z g z t ẋ2 + ẏ 2 + ż 2 ) = 2ẋẍ + 2ẏÿ + 2ż z = 2λ m ẋ g x + ẏ g y + ż g ) = 0 z an so ẋ 2 +ẏ 2 +ż 2 s constant an so s t s constant so changng from t to s gves the geoesc equaton. Hence the moton of a free partcle s along a geoesc an constant spee. 20 of 20