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 xxis. 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 xyplne 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: EulerLgrnge 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: EulerLgrnge 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 EulerLgrnge eqution, Not every function f (x) which stis es the EulerLgrnge 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 EulerLgrnge 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 EulerLgrnge 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 EulerLgrnge = 0 since, in fct, F is independent of x: Extensions of the EulerLgrnge 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 ndimensionl (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 EulerLgrnge 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 ndimensionl (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 EulerLgrnge 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 EulerLgrnge 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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