Integrals along Curves.


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1 Integrls long Curves. 1. Pth integrls. Let : [, b] R n be continuous function nd let be the imge ([, b]) of. We refer to both nd s curve. If we need to distinguish between the two we cll the function the prmetriztion of the curve nd the imge (or the pth) of the curve. The point () is clled the initil (end) point nd (b) is clled the finl (end) point. There re such continuous functions whose imge do not t ll look like wht we think of curve. For exmple, there re spcefilling curves in the plne whose imges contin solid rectngle. Such curves hve, intuitively speking, n infinite length. So our first restriction will be: Definition. A curve : [, b] R n is sid to be rectifible if there exists number K such tht for every prtition P of the domin, P : = t 0 < t 1 < < t 1 we hve l P := (t j ) (t j 1 ) K 2. If curve is rectifible then we define its length to be l() := sup{l P : P is prtition of [, b] }. Definition We sy tht curve : [, b] R n is smooth curve if is continuously differentible 3. We sy tht it is piecewise smooth curve if there exists finite prtition P : = t 0 < t 1 < < t N such tht the restrictions of to the intervls [t i 1, t i ] re ll smooth curves. Using this ide we my, more generlly define the integrl of bounded function f : R long curve. For given prtition P let τ i [t i 1, t i ] ( mrking of the prtition). Let P := mx i t i t i 1. We my define the Riemnn sums nd the Riemnn integrl s limit of such sums: f(x) ds := lim f((τ j )) (t j ) (t j 1 ) P 0 whenever this limit exists. It cn be shown tht this limit exists if, for exmple, f is continuous. We could lterntively use upper nd lower sums to define integrbility, but show tht both methods yield the sme result. All this is good to know for your generl knowledge of mthemtics, however we will usully sty wy from such generlity nd restrict ourselves to piecewise smooth curves. If the curve is smooth then (see the theorem below): b f(x) ds = f((t)) (t) dt, with the obvious generliztion to integrls long piecewise smooth curves s sums of such integrls long smooth curves. 1 N = since the number of prtition points depends on the prtition. 2 x y denotes ordinry Eucliden distnce between the points x nd y nd so l P is the length of the polygonl line connecting the points (t j) 3 At the endpoints we, of course, interpret differentible in the onesided sense. 1
2 These ides trnslte in n obvious wy to curves in C nd integrls of complex vlued functions long curves. Definition. A curve : [, b] C is sid to be rectifible if there exists number K such tht for every prtition P of the domin, P : = t 0 < t 1 < < t we hve l P := (t j ) (t j 1 ) K. If curve is rectifible then we define its length to be l() := sup{l P : P is prtition of [, b] }. Definition We sy tht curve : [, b] C is smooth curve if is continuously differentible nd is nowhere zero. We sy tht it is piecewise smooth curve if there exists finite prtition P : = t 0 < t 1 < < t N such tht the restrictions of to the intervls [t i 1, t i ] re ll smooth curves. Suppose tht f : C is bounded function defined long curve. For given prtition P let τ i [t i 1, t i ] be mrking of the prtition. Let P := mx i t i t i 1. We my define f(z) ds := lim f((τ j )) (t j ) (t j 1 ) P 0 whenever this limit exists. Of course, if we write f in terms of its rel nd imginry prts, f = g+ih, then f(z) ds = g(x + iy) ds + i h(x + iy) ds. If the curve is smooth then (see the theorem below): f(x) ds = b f((t)) (t) dt, with the obvious generliztion to integrls long piecewise smooth curves. Definitions. We sy tht the curves : [, b] R n nd β : [c, d] R n re equivlent, nd we write δ, if there exists continuous, strictly incresing function µ : [, b] [c, d] such tht = β µ. If we let [] be the equivlence clss of curve, then we let [] be the equivlence clss of β where β : [, b] R n defined by β(t) = ( + b t). Tht is, the sme curve ( = β ) but with the opposite orienttion. If α : [p, q] R n nd : [, b] R n re two curves such tht the terminl point of α is the initil point of then we cn define [β] = [α] + [] in nturl wy, for exmple { α(t) if p t q β(t) = (t + q) if q t q + b, 2
3 so tht the pth of β is the pth of α, followed by the pth of. If the initil point of curve is lso its terminl point then we sy the curve is closed. If the curve does not intersect itself, except possibly tht the initil point is equl to the terminl point, then we sy tht the curve is simple. Definition Suppose tht the curve : [, b] R n is differentible t t 0 nd tht (t 0 ) 0. Any vector of the form c (t 0 ), c 0, is clled tngent vector t the point (t 0 ). The line x = (t 0 ) + s (t 0 ), s R is clled the tngent line to the curve t the point (t 0 ). The plne (or line, or hyperplne, depending on the dimension) through this point nd norml to (t 0 ) is clled the norml plne. It hs the eqution (t 0 ) (x (t 0 )) = 0. Theorem 1 If : [, b] R n is smooth curve nd f is continuous function on then l() = f(x) ds = b b (t) dt, f((t)) (t) dt. The proof is lengthy (pp [1]) but n excellent exmple of proof in nlysis, nd you re encourged to red it. Note tht integrls long equivlent curves hve the sme vlue, i.e. the integrl does not depend on the prmeteriztion. This is simple consequence of the chin rule. 2. RiemnnStieltjes integrls. There is simple generliztion of the Riemnn integrl, clled the RiemnnStieltjes integrl. The RiemnnStieltjes integrl of relvlued function f of rel vrible with respect to rel function g is denoted by b f(t)dg(t) nd defined to be the limit, s the mesh of the prtition P := { = t 0 < t 1 < < t N = b} of the intervl [, b] pproches zero, of the pproximting sum N S(P, f, g) := f(τ i )[g(t i ) g(t i 1 )] i=1 where τ i [t i 1, t i ]. The two functions f nd g re respectively clled the integrnd nd the integrtor. The limit is here understood to be number A (the vlue of the RiemnnStieltjes integrl) such tht for every ɛ > 0 there exists δ > 0 such tht for every prtition P with mesh P < δ, nd for every choice of points τ i [t i 1, t i ], we hve S(P, f, g) A < ɛ. If g is nondecresing function on [, b], then the generlized RiemnnStieltjes of with respect to g exists if nd only if, for every ɛ > 0, there exists prtition P such tht U(P, f, g) L(P, f, g) < ɛ (the Riemnn lemm). If g is continuously differentible then this integrl my be reduced to n ordinry Riemnn integrl: b f(t)dg(t) = b f(t)g (t)dt. 3
4 Exercise. Prove the bove ssertion. However, g my hve jump discontinuities, or my hve derivtive zero lmost everywhere while still being continuous nd incresing, in either of which cses the RiemnnStieltjes integrl is not cptured by ny expression involving derivtives of g. Exmple. Let f(t) = 1/t 2 nd g(t) = t, i.e. g(t) is the lrgest integer tht is less thn or equl to t. Then for ny positive integer N 3. Line Integrls. N 0 f(t)dg(t) := N n=1 1 n 2. Definition. Let : [, b] U R n be rectifible curve, x i (t) = i (t), i = 1, 2,, n nd let f : U R n, f = (f 1, f 2,, f n ), be piecewise continuous function, then for 1 k n we define f k dx k := b f k ((t))d k (t) f k (x)dx k nd f dx = n i=1 f i (x) dx i. Exercise. Show tht if the curve is piecewise smooth then f dx = n i=1 b f i ((t)) dx i dt dt. From now on we will only del with curves tht re smooth or t lest piecewise smooth. Properties. f dx = f dx δ+ f dx = δ f dx + f dx. Definitions. We sy tht f is grdient field in D R n if φ C 1 (D) such tht f = φ in D. In this cse φ is sid to be potentil function for f. If for ech : [, b] D tht is piecewise smooth the vlue of the line integrl f dx depends only on the endpoints then we sy tht line integrls of f re pth independent. In this cse we sy tht the field f is conservtive. If f is grdient field then the differentil form f 1 dx 1 + f 2 dx f n dx n is sid to be exct. 4
5 Theorem 2. Suppose tht D is domin in R n nd tht f C(D, R n ). then the following re equivlent: f is grdient field. f is conservtive. For ny piecewise smooth closed curve we hve f dx = 0. The mjor prt of the proof is to show tht if f is conservtive then given ny point x 0 D, φ(x) := x x 0 f dx long ny piecewise smooth curve from x 0 to x is well defined nd is potentil for f. Definition. A set S R n is sid to be convex if the line segment uv S whenever u, v S. Let us consider the twodimensionl sitution. If f(x.y) := (P (x, y), Q(x, y)) is continuously differentible grdient field defined on some open set D, then there exists (twice continuously differentible) potentil function φ such tht P = φ x nd Q := φ y. Therefore P y = φ xy = φ yx = Q x 4. In generl the differentil form P (x, y)dx + Q(x, y)dy is sid to be closed if P y = Q x. Hence P (x, y)dx+q(x, y)dy is closed if it is exct. The converse is not true nd is, in fct, highly dependent on the geometry of the domin D. A creful study of closed versus exct leds us to the field of Algebric Topology. Green s Theorem. Let Ω R 2 be convex nd bounded, nd suppose tht its boundry, Ω, is piecewise smooth. Suppose tht P nd Q re continuously differentible on n open set U Ω Ω. Then ( Q P dx + Qdy = x P ) dx dy, y D where the line integrl is tken in the counterclockwise sense 5. Ω Corollry 1. Let Ω R 2 be convex nd bounded, nd suppose tht its boundry, Ω, is piecewise smooth. Suppose tht P nd Q re continuously differentible on n open set U Ω Ω. Suppose lso tht P y = Q x in Ω. Then P dx + Qdy = 0. D 4 Recll tht for twice continuously differentible functions the mixed second order derivtives re independent of the order in which the derivtives re tken. 5 The requirement tht Ω be convex cn be gretly relxed. For exmple it is sufficient to ssume tht Ω is simply connected. 5
6 4. Complex Line Integrls. Definition Let D be n open set in C nd f : D C bounded function. Suppose tht : [, b] D is rectifible curve in D. Let P := {(t j ) = t 0 < t 1 < < t = b} denote generic prtition of the curve nd let τ j [t j 1, t j ] denote mrking of the prtition. We define f(z) dz := lim f((τ j ))((t j ) (t j 1 )) P 0 whenever this limit exists (irrespective of the choice of the mrking). It cn be shown tht this limit exists if, for exmple, f is continuous. Letting z j := (t j ) nd ζ j := (t j ) we hve the more compct eqution f(z) dz := lim f(ζ j )(z j z j 1 ). P 0 Note tht this is simply complex RiemnnStieltjes integrl. To simplify mtters we will restrict ourselves to piecewise smooth curves tht re mde up of finitely mny sections of smooth curve. We my write f(z) = u(z) + iv(z). If is smooth curve then (t) = 1 (t) + i 2 (t) nd there exist, by the Men Vlue Theorem, τ j, σ j [t j 1, t j ] so tht z j z j 1 = [ 1 (σ j) + i 2 (τ j)](t j t j 1 ). This mens tht we cn write the Riemnn sums [u((τ j )) 1(τ j ) + iu((σ j )) 2(σ j ) + iv((τ j )) 1(τ j ) v((σ j )) 2(τ j )](t j t j 1 ). Letting P tend to zero we see tht f(z) dz = b [u((t) 1(t) + iu((t)) 2(t) + iv((t)) 1(t) v((t)) 2(t)] dt = So we hve the sme formul s the definition of line integrl in the text, [2]: f(z) dz = b b f((t)) (t) dt. f((t)) (t) dt. (1) 5. Properties of Line Integrls in C. Theorem 3 Suppose f nd g re integrble long the curves nd δ, nd let, b C then () (b) [f(z) + bg(z)] dz = f(z) dz + b g(z) dz. f(z) dz = f(z) dz. 6
7 (c) (d) +δ f(z) dz f(z) dz = f(z) dz + f(z) dz. δ f(z) ds Ml(), where M := sup f(z). The proofs of these, for the cse where the integrnds re continuous nd where the pths re smooth, re given in the text [2], but they esily follow from the definition of line integrls of integrble functions long rectifible curves. Theorem 4. Suppose f is differentible on the curve : [α, β] C then f (z) dz = f((β)) f((α)). Corollry 2. Suppose f hs continuous ntiderivtive F, i.e. F (z) = f(z), on the simple closed curve : [α, β] C then f(z) dz = 0. Theorem 5. Let Ω C be convex, open, nd bounded, nd suppose tht its boundry, Ω, is smooth. Suppose tht f re continuously differentible on n open set U Ω Ω. Then f(z) dz = 0. (2) D The proof follows immeditely from Green s theorem nd the CuchyRiemnn equtions (esy exercise). Note the difference in the sttements of theorem 5 nd corollry 2 6. To illustrte the difference consider the following exmple Exmple. Let B be the unit disk centered t the origin. Then B =, the unit circle. We orient counterclockwise. We cn show tht 1 dz = 2πi. z Note tht 1/z is not differentible on ll of B. Corollry 2 implies tht there pprently does not exist n ntiderivtive for 1/z on ll of. We will understnd this more when we study the (multivlued) function ln(z). own. 6 The text [2] does not use Green s Theorem nd so the proof there is lengthier, but hs the dvntge of stnding on its 7
8 Note. The strtegy for the proof in the text [2] is not to use Green s Theorem but to follow the following steps: () Consider the cse of stndrd rectngles in C, tht is to sy rectngles whose sides re prllel to the xes, nd to initilly only consider entire functions f, i.e. functions tht re nlytic everywhere in C. (b) Show tht f(z) dz = 0 if is stndrd rectngle. (c) Prove tht if f is entire then we cn construct n nlytic function F such tht F = f. (d) To show tht f(z) dz = 0 for ny piecewise smooth simple closed curve. References [1] Friedmn, Avner. Advnced Clculus, Dover Publictions. [2] Bk, Joseph nd Donld Newmn. Complex Anlysis, 3rd edition, Springer. 8
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