4. Calculus of Variations

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1 4. Clculus of Vritions Introduction - Typicl Problems The clculus of vritions generlises the theory of mxim nd minim. Exmple (): Shortest distnce between two points. On given surfce (e.g. plne), nd the shortest curve between two points. The length L of curve y = y (x) between the vlues x = nd x = b is given by the integrl L = s 1 + dy dx 2 dx = q 1 + (y 0 ) 2 dx: (4.1) The vlue of L depends upon the function y (x) which ppers s n rgument in the integrnd, n rbitrry continuous function with piecewise continuous drivtive. Exmple (b) Miniml surfce of revolution. Let the curve y = y (x) 0 which psses through the points y () = y 1, y (b) = y 2 be rotted bout the x-xis. The resulting surfce between x = nd x = b hs surfce re A given by q A = 2 y 1 + (y 0 ) 2 dx: The curve y = y (x) which gives the smllest surfce of revolution is found by minimising the integrl. Exmple (c): Isoperimetric problem. Find closed plne curve of given perimeter which encloses the gretest re. The re my be written s A = Z between some limits (nd tking cre bout regions bove nd below the x- xis) nd is subject to constrint of the form equ. (4.1) where now L (the perimeter) is xed in length. Common ingredients: (1) An integrl with n integrnd contining n rbitrry function (2) A problem which sks for minimum or mximum. Note the geometric lnguge nd tht the concepts of "curve" nd "function" do not coincide. Functionls Let S be (vector) spce (i.e. set of functions + lgebric structure - closed under ddition of functions f + g nd multipliction by sclr, f): Exmple: Let C 1 (; b) denote the set of functions continuous on the closed intervl [; b] with piecewise continuous rst order derivtives. ydx 1

2 4.2 De nition: A functionl is mpping (function) from spce of functions into the underlying eld (usully the rel or complex numbers) : S! R(or C) : y 7! c y 2 S; c 2 R( C) (y) = c; or y = c: S is clled the domin of the functionl nd the spce of dmissible functions. Exmples (): Evluting function is functionl, e.g. for d 2 [; b] (y (x)) = y (c) (b) (y (x)) = y 00 (7) + y (3) : (c) The Dirc delt "function" is functionl: (x d) f (x) dx = f (d) (d) Let y (x) 2 C 1 (; b) ; (y (x)) = h (y (x)) 2 (y 0 (x)) 2i dx: (e) The re A of surfce z = z (x; y) lying bove the region G in the xy-plne is given by ZZ q A = 1 + zx 2 + zydxdy 2 G where z x = z y nd is functionl of the rgument function z (x; y). The clculus of vritions is concerned with nding extrem or sttionry vlues of functionls. Consider functionls (de ned by integrls) of the form I (y) = F (x; y; y 0 ) dx: (4.3) The integrnd F depends on the function y (x), its derivtive y 0 (x) nd the independent vrible x: In order to discuss mxim nd minim we need to de ne wht is ment by two functions being "close together", i.e. we need notion of distnce. 4.4 De nition: Given h 2 R, h > 0; function y 1 (x) lies in the neighbourhood N h (y) of the function y (x) if jy (x) y 1 (x)j < h 2

3 8x 2 [; b] : Sometimes it is necessry to use more re ned de nition: 4.4 De nition: Given h 2 R, h > 0; function y 1 (x) lies in the rst order neighbourhood N h (y) of the function y (x) if jy (x) y 1 (x)j < h nd jy 0 (x) y 0 1 (x)j < h 8x 2 [; b] : 4.5 Fundmentl problem of the Clculus of Vritions: Find function y = y 0 (x) 2 S (; b) for which the functionl I (y) tkes n extreml vlue (i.e. mximum or minimum) vlue with respect to ll y(x) 2 S (; b) in N h (y) for su ciently smll h. y = y 0 (x) is clled n extreml function. Note: There is no gurntee solution exists for this problem (unlike mxim nd minim of functions continuous on closed intervl where existence is gurnteed). Exmple: The shortest distnce between two points A; B is stright line but there is no curve of shortest length which deprts from A nd rrives t B t right ngles to the line segment AB: Euler-Lgrnge Equtions 4.6 Fundmentl lemm in the Clculus of Vritions Let f (x) be continuous in [; b] nd let (x) be n rbitrry function on [; b] such tht ; 0 ; 00 re continuous nd () = (b) = 0: If f (x) (x) dx = 0 for ll such (x) then f (x) 0 on [; b] : Proof: Suppose to the contrry w.l.o.g. tht f (x) > 0 t, sy, x = : Then there is neighbourhood N, 0 < x < 1 in which f (x) > 0: Let (x 0 ) 4 (x (x) = 1 ) 4 for x 2 N 0 elsewhere : Then f (x) (x) dx > 0 contrdicting the hypothesis. 4.7 Theorem: Euler-Lgrnge Equtions The extreml function y = y 0 (x) for the functionl (4.3) I (y) = F (x; y; y 0 ) dx 3

4 where ; b; y () ; y (b) re given, F is twice continuously di erentible w.r.t. its rguments nd y 00 (x) is continuous, stis es the eqution d dx 0 = 0: This is necessry, but not su cient, condition for n extreml function. (Every extreml function y 0 (x) stis es the Euler-Lgrnge eqution, Not every function f (x) which stis es the Euler-Lgrnge eqution is n extreml function.) Proof: Let y = y (x) be vrible function nd let y = y 0 (x) be n extreml function for the functionl I (y), i.e. I (y 0 ) tkes n extreme vlue (mximum or minimum). Let (x) be s in lemm (4.6) nd de ne the function where " > 0 is prmeter, nd write y (x) = y 0 (x) + " (x) y = y 0 + y: Then y = " (x) is clled the vrition of y = y 0 (x). For " su ciently smll, y lies in n rbitrrily smll neighbourhood N h (y 0 ) of y 0 (x). Now, the integrl I (y) = I (y 0 + ") is function of (") of ": Let I = I (y 0 + ") I (y 0 ) = (") (0) then I = [F (x; y 0 + "; y " 0 ) F (x; y 0 ; y 0 0)] dx: Expnding (") in Mclurin series with respect to " to rst order gives nd so to rst order in " so I = 0 (0) " (") = (0) + 0 (0) " 0 0 (0) = dy d" + dy 0 0 dx d" dx: y=y0 Now, since y = y 0 (x) + " (x) where y 0 (x) is the extreml function, it follows tht I = 0 for ll (x) with y 2 N h (y 0 ) : For suppose not, then replce (x) by (x) nd the sign of I chnges, contrdicting the fct tht y 0 (x) is the extreml function! 1 A " 4

5 Hence I (y) tkes sttionry vlue for y = y 0 (x) : Thus dx = 0: Integrting the second integrl by prts with then gives b 0 + Now, () = (b) = 0 so u = 0 ; v0 = 0 u 0 = d dx 0 ; v = d dx 0 dx = 0: d dx 0 dx = 0: Hence, since (x) is rbitrry, using the Fundmentl Lemm of the Clculus of Vritions, we obtin d dx 0 = 0: Note: For ny such f (x; y; y 0 ) so putting f = = 0 gives df dy dx dy 0 0 dx This is 2nd. order 2 F 02 y00 (x) F 0 y0 (x) 0 = 0: Exmple: Find the extreml function for with y (0) = 0; y 1 2 = 1: Here, so I (y) = Z =2 0 h (y 0 ) 2 y 2i dx F (x; y; y 0 ) = (y 0 ) 2 y 2 = 2y; 0 = 2y0 : 5

6 The Euler-Lgrnge eqution d dx 0 = 0 becomes i.e. The generl solution is y (0) = 0 ) A = 0 nd y d dx (2y0 ) ( 2y) = 0 y 00 + y = 0: y (x) = A cos x + B sin x 1 2 = 1 ) B = 1 hence y = sin x nd I (sin x) is sttionry nd I (sin x) = Z =2 0 cos 2 x sin 2 x dx = 0: Specil cses: First integrls of the Euler-Lgrnge equtions For specil forms of F (x; y; y 0 ) the 2nd. order order ode d dx 0 = 0 my be integrted to give " rst integrl". 4.8 Theorem: Let F F (x; y 0 ) (no y dependence) then Proof: so 0 = constnt. = 0 d dx 0 = 0 which integrtes w.r.t. x to give the result. 4.9 Theorem: Let F F (y; y 0 ) (no x dependence) then F y 0 0 = constnt. 6

7 Proof: In generl F F (x; y; y 0 ) d F y 0 dx 0 + y0 + + y0 0 y00 d dx 0 00 y 0 = by the Euler-Lgrnge = 0 since, in fct, F is independent of x: Extensions of the Euler-Lgrnge equtions () Severl unknown functions y 1 (x) ; :::; y m (x) : Let y 0 d dx 0 then I (y 1 ; :::; y m ) = d dx F (x; y 1 ; :::; y m ; y 0 1; :::; y 0 m) dx (4.10)! 0 j j = 0; (4.11) where j = 1; :::; m; i.e. m simultneous ode s for the m unknowns y j : (b) Severl independent vribles x 1 ; :::; x n : Let y = y (x 1 ; :::; x n ) nd Z Z I (y) = ::: F x 1 ; :::; x n ; y; ; :::; dx 1 :::dx n (4.12) n ( multiple integrl) where V is region in n-dimensionl (x 1 ; :::; x n )-spce then, writing y ;xi y ;i y i we hve i ;i = 0; (4.13) prtil di erentil eqution. (c) Additionl integrl constrint - Lgrnge multipliers Suppose tht, in ddition to Eq. (4.3), y (x) lso stis es n integrl constrint of the form J (y) = G (x; y; y 0 ) dx = C (4.14) where C is constnt. We wish to nd sttionry vlues of the functionl I (y) = F (x; y; y 0 ) dx 7

8 where y (x) is now not free, but is subject to the constrint Equ.(4.13). In this cse, form the functionl K (y) = = (F G) dx H (x; y; y 0 ; ) dx, sy where is Lgrnge multiplier. The corresponding Euler-Lgrnge eqution dx 0 = 0: (4.15) The generl solution of equ. (4.15) contins nd two integrtion constnts. These re determined by the boundry conditions nd the constrint equ. (4.14). Given severl constrints where k = 1; :::; p; de ne J k (y) = H = F G k (x; y; y 0 ) dx = C k px k G k : Imposing the constrints gives the vlues of k (usully of little physicl significnce). (d) Severl functions of severl vribles. Now consider m functions y 1 ; :::y m ech of n vribles x 1 ; :::x n ; where j = 1; :::; m nd i = 1; :::; n: Let k=1 y j = y j (x 1 ; :::x n ) = y j (x i ) ; y ji = i nd let V denote region in the n-dimensionl (x 1 ; :::x n )-spce, dv = dx 1 :::dx n : Consider the functionl Z I (y 1 ; :::; y m ) = F x 1 ; :::; x n ; y 1 ; :::; y m ; 1 ; :::; m dx 1 :::dx n n which my be bbrevited by Z I (y j ) = V V F (x i ; y j ; y ji ) dx 1 :::dx n : The Euler-Lgrnge equtions re the m equtions 8

9 i ji j = 0; (4.17) where j = 1; :::; m. Note tht, using the double su x summtion convention (dssc) we my bbrevite this = i ji j where, ny term contining repeted subscript (here i) is summed over ll possible vlues (here i tkes the vlues 1; :::; n). A single subscript occurring in every term of n eqution (here j) is cllled free subscript nd indictes tht there is n eqution for ech vlue (here m). Exmple: Dirichlet s integrl nd Poisson s eqution. The bove nottion ws developed s generlistion of of the originl nottion for the simplest cse, the functionl (4.3) nd the originl Euler-Lgrnge eqution in theorem (4.7). It is not prticulrly convenient for considering single prtil di erentil eqution. So, we now chnge nottion. Let n = 3 nd let the independent vribles x 1 ; x 2 ; x 3 be replced by x; y; z (more suitble for problems in three spce dimensions). Also replce the dependent vrible y (now being used s coordinte) by u. Thus, let u = u (x; y; z) be function of three vribles. Recll the nottion for the grdient ru @z nd the nottion (ru) 2 = ru ru = + Now, Dirichlet s integrl is de ned by Z 1 I (u) = 2 (ru)2 + fu dv V = u 2 x + u 2 y + u x: where V is volume in R 3 ; f = f (x; y; z) is known function of x; y; z. Here, F = = x = u x y = u y ; The Euler Lgrnge equtions y z = u z = 0 9

10 i.e. or where r 2 is the (u x) (u (u z) f = 0 r 2 u = f r : Thus, the Dirichlet integrl is sttionry when the function u (x; y; z) stis es Poisson s eqution. This is known s Dirichlet s Principle. This is simple exmple of vritionl principle, which turns out to be very importnt ppliction of the clculus of vritions. 10

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