One dimensional integrals in several variables

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Chpter 9 One dimensionl integrls in severl vribles 9.1 Differentition under the integrl Note: less thn 1 lecture Let f (x,y be function of two vribles nd define g(y : b f (x,y dx Suppose tht f is differentible in y. The question we sk is when cn we simply differentite under the integrl, tht is, when is it true tht g is differentible nd its derivtive g (y? b f (x,y dx. y Differentition is limit nd therefore we re relly sking when do the two limitting opertions of integrtion nd differentition commute. As we hve seen, this is not lwys possible. In prticulr, the first question we would fce is the integrbility of f y. Let us prove simple, but the most useful version of this theorem. Theorem 9.1.1 (Leibniz integrl rule. Suppose f : [,b] [c,d] R is continuous function, such exists for ll (x,y [,b] [c,d] nd is continuous. Define tht f y Then g: [c, d] R is differentible nd g(y : g (y b b f (x,y dx. f (x,y dx. y 53

54 CHAPTER 9. ONE DIMENSIONAL INTEGRALS IN SEVERAL VARIABLES Note tht the continuity requirements for f nd f y cn be wekened but not dropped outright. The min point is for f y to exist nd be continuous for smll intervl in the y direction. In pplictions, the [c,d] cn be mde very smll intervl round the point where you need to differentite. Proof. Fix y [c,d] nd let ε > be given. As f y is continuous on [,b] [c,d] it is uniformly continuous. In prticulr, there exists δ > such tht whenever y 1 [c,d] with y 1 y < δ nd ll x [,b] we hve f y (x,y 1 f y (x,y < ε. Now suppose h is such tht y + h [c,d] nd h < δ. Fix x for moment nd pply men vlue theorem to find y 1 between y nd y + h such tht If h < δ then f (x,y + h f (x,y h f (x,y + h f (x,y h f y (x,y 1. f y (x,y f y (x,y 1 f y (x,y < ε. This rgument worked for every x [,b]. Therefore, s function of x x f (x,y + h f (x,y h converges uniformly to x f (x,y s h. y We only defined uniform convergence for sequences lthough the ide is the sme. If you wish you cn replce h with 1/n bove nd let n. Now consider the difference quotient g(y + h g(y h b f (x,y + h dx b f (x,y dx h Uniform convergence cn be tken underneth the integrl nd therefore g(y + h g(y lim h h b lim h f (x,y + h f (x,y h b dx f (x,y + h f (x,y h b f (x,y dx. y dx. Exmple 9.1.2: Let Then f (y f (y 1 1 sin(x 2 y 2 dx. 2ycos(x 2 y 2 dx.

9.1. DIFFERENTIATION UNDER THE INTEGRAL 55 Exmple 9.1.3: Suppose we strt with 1 x 1 ln(x dx. The function under the integrl cn be checked to be continuous, nd in fct extends to be continuous on [,1], nd hence the integrl exists. Trouble is finding it. Introduce prmeter y nd define function. 1 x y 1 f (y : ln(x dx. Agin it cn be checked tht xy 1 ln(x is continuous function (tht is extends to be continuous of x nd y for (x, y [, 1] [, 1]. Therefore f is continuous function of on [, 1]. In prticulr f (. For ny ε >, the y derivtive of the integrnd, tht is x y is continuous on [,1] [ε,1]. Therefore, for y > we cn differentite under the integrl sign f (y 1 ln(xx y 1 ln(x dx x y dx 1 y + 1. We need to figure out f (1, knowing f (y y+1 1 nd f (. By elementry clculus we find f (1 1 f (y dy ln(2. Therefore 1 x 1 dx ln(2. ln(x Exercise 9.1.1: Prove the two sttements tht were sserted in the exmple. Prove x 1 ln(x extends to continuous function of [,1]. b Prove xy 1 ln(x extends to be continuous function on [,1] [,1]. 9.1.1 Exercises Exercise 9.1.2: Suppose h: R R is continuous function. Suppose tht g: R R is which is continuously differentible nd compctly supported. Tht is there exists some M > such tht g(x whenever x M. Define f (x : h(yg(x y dy. Show tht f is differentible. Exercise 9.1.3: Suppose f is n infinitely differentible function (ll derivtives exist such tht f (. Then show tht there exists nother infinitely differentible function g(x such tht f (x xg(x. Finlly show tht if f ( then g(. Hint: first write f (x x f (sds nd then rewrite the integrl to go from to 1.

56 CHAPTER 9. ONE DIMENSIONAL INTEGRALS IN SEVERAL VARIABLES Exercise 9.1.4: Compute 1 etx dx. Derive the formul for 1 xn e x dx not using itnegrtion by prts, but by differentition underneth the integrl. Exercise 9.1.5: Let U R n be n open set nd suppose f (x,y 1,y 2,...,y n is continuous function defined on [,1] U R n+1. Suppose f y 1, f y 2,..., f y n exist nd re continuous on [,1] U. Then prove tht F : U R defined by 1 F(y 1,y 2,...,y n : f (x,y 1,y 2,...,y n dx is continuously differentible.

9.2. PATH INTEGRALS 57 9.2 Pth integrls Note:??? lectures 9.2.1 Piecewise smooth pths Definition 9.2.1. A continuously differentible function : [,b] R n is clled smooth pth or continuously differentible pth if is continuously differentible nd (t for ll t [,b]. The function is clled piecewise smooth pth or piecewise continuously differentible pth if there exist finitely mny points t < t 1 < t 2 < < t k b such tht the restriction of the function [t j 1,t j ] is smooth pth. We sy is simple pth if (,b is one-to-one function. A is closed pth if ( (b, tht is if the pth strts nd ends in the sme point. Since is function of one vrible, we hve seen before tht treting (t s mtrix is equivlent to treting it s vector since it is n n 1 mtrix, tht is, column vector. In fct, by n exercise, even the opertor norm of (t is equl to the eucliden norm. Therefore, we will write (t s vector s is usul. Generlly, it is the direct imge ( [,b] tht is wht we re interested in. We will informlly tlk bout curve, by which we will generlly men the set ( [,b]. Exmple 9.2.2: Let : [,4] R 2 be defined by (t, if t [,1], (1,t 1 if t (1,2], (t : (3 t,1 if t (2,3], (,4 t if t (3,4]. Then the reder cn check tht the pth is the unit squre trversed counterclockwise. We cn check tht for exmple [1,2] (t (1,t 1 nd therefore ( [1,2] (t (,1. It is good to notice t this point tht ( [1,2] (1 (,1, ( [,1] (1 (1,, nd (1 does not exist. Tht is, t the corners is of course not differentible, even though the restrictions re differentible nd the derivtive depends on which restriction you tke. Exmple 9.2.3: The condition tht (t mens tht the imge of hs no corners where is smooth. For exmple, the function { (t 2, if t <, (t : (,t 2 if t. Note tht the word smooth is used sometimes of continuously differentible, sometimes for infinitely differentible in the literture.

58 CHAPTER 9. ONE DIMENSIONAL INTEGRALS IN SEVERAL VARIABLES It is left for the reder to check tht is continuously differentible, yet the imge (R {(x,y R 2 : (x,y ( s, or (x,y (,s for some s } hs corner t the origin. Exmple 9.2.4: A grph of continuously differentible function f : [,b] R is smooth pth. Tht is, define : [,b] R 2 by (t : ( t, f (t. Then (t ( 1, f (t, which is never zero. There re other wys of prmetrizing the pth. Tht is, hving different pth with the sme imge. For exmple, the function tht tkes t to (1 t +tb, tkes the intervl [,1] to [,b]. So let α : [,1] R 2 be defined by α(t : ( (1 t +tb, f ((1 t +tb. Then α (t ( b,(b f ((1 t+tb, which is never zero. Furthermore s sets α ( [,1] ( [,b] {(x,y R 2 : x [,b] nd f (x y}, which is just the grph of f. The lst exmple leds us to definition. Definition 9.2.5. Let : [,b] R n be piecewise smooth pth nd h: [c,d] [,b] continuously differentible function such tht h (t for ll t [c,d]. Then the composition h is clled smooth reprmetriztion of. If h (t < for t [c,d], then h is sid to reverses orienttion. If h does not reverse orienttion then h is sid to preserve orienttion. A reprmetriztion is nother pth for the sme set. Tht is, ( h ( [c,d] ( [,b]. Let us remrk tht since the function h is continuous nd h (t for ll t [c,d], then if h (t < for one t [c,d], then h (t < for ll t [c,d] by the intermedite vlue theorem. Tht is, h (t hs the sme sign t every t [c,d]. Proposition 9.2.6. If : [,b] R n is piecewise smooth pth, nd h: [c,d] R n is smooth reprmetriztion, then h is piecewise smooth pth. Proof. If h: [c,d] [,b] gives smooth reprmetriztion, then s h (t hs the sme sign for ll t [c,d]. Hence it is bijective mpping with continuously differentible inverse. Suppose t < t 1 < t 2 < < t k b is the prtition from the definition of piecewise smooth for. Suppose first tht h preserves orienttion, tht is h is strictly incresing. Let s j : h 1 (t j. Then s c < s 1 < s 2 < < s k d. For t [s j 1,s j ] notice tht h(t [t j 1,t j ] nd so ( h [s j 1,s j ](t [t j 1,t j ]( h(t. The function ( h [s j 1,s j ] is therefore continuously differentible nd by the chin rule ( (t ( ( ( h [s j 1,s j ] [t j 1,t j ] h(t h (t. Therefore h is piecewise smooth pth. One need not hve the reprmetriztion be smooth t ll points, it relly only needs to be gin piecewise smooth, but the bove definition will suffice for us nd it keeps mtters simpler.

9.2. PATH INTEGRALS 59 9.2.2 Pth integrl of one-form Definition 9.2.7. If (x 1,x 2,...,x n R n re our coordintes, nd given n rel-vlued continuous functions f 1, f 2,..., f n defined on some set S R n we define so clled one-form: ω f 1 dx 1 + f 2 dx 2 + f n dx n. Tht is, we could represent ω s function from S to R n. Exmple 9.2.8: For exmple, ω(x,y is one-form defined on R 2 \ {(,}. y x 2 + y 2 dx + x x 2 + y 2 dy Definition 9.2.9. Let : [,b] R n be smooth pth nd ω f 1 dx 1 + f 2 dx 2 + f n dx n, one-form defined on the direct imge ( [,b]. Let ( 1, 2,..., n be the components of. Then define b ( ( ( ( ω : f 1 (t 1 (t + f 2 (t 2 (t + + f n (t n (t dt b ( n j1 f j ( (t j (t dt. If is piecewise smooth with the corresponding prtition t < t 1 < t 2 <... < t k b, then ech [t j 1,t j ] is smooth pth nd we define ω : ω + ω + + ω. [t,t 1 ] [t1,t 2 ] [tn 1,tn] The nottion mkes sense from the formul you remember from clculus, let us stte it somewht informlly: if x j (t j (t, then dx j j (tdt. Exmple 9.2.1: Let the one-form ω nd the pth : [,2π] R 2 be defined by Then ω 2π 2π ω(x,y : ( y x 2 + y 2 dx + x x 2 + y 2 dy, (t : ( cos(t,sin(t. sin(t ( cos(t ( ( 2 ( 2 sin(t + ( 2 ( 2 cos(t dt cos(t + sin(t cos(t + sin(t 1dt 2π.

6 CHAPTER 9. ONE DIMENSIONAL INTEGRALS IN SEVERAL VARIABLES Next, let us prmetrize the sme curve s α : [,1] R 2 defined by α(t : ( cos(2πt,sin(2πt, tht is α is smooth reprmetriztion of. Then ( 1 sin(2πt ( ω ( 2 ( 2 2π sin(2πt cos(2πt + sin(2πt α 1 + cos(2πt ( cos(2πt 2 + ( sin(2πt 2 ( 2π cos(2πt dt 2π dt 2π. Now let us reprmetrize with : [,2π] R 2 s (t : ( cos( t,sin( t. Then ( 2π sin( t ( cos( t ( ω ( 2 ( 2 sin( t + ( 2 ( 2 cos( t dt cos( t + sin( t cos( t + sin( t 2π ( 1dt 2π. Now, α ws n orienttion preserving reprmetriztion of, nd the integrl ws the sme. On the other hnd is n orienttion reversing reprmetriztion nd the integrl ws minus the originl. The previous exmple is not fluke. The pth integrl does not depend on the prmetriztion of the curve, the only thing tht mtters is the direction in which the curve is trversed. Proposition 9.2.11. Let : [,b] R n be piecewise smooth pth nd h: [c,d] R n smooth reprmetriztion. Suppose ω is one-form defined on the set ( [,b]. Then { ω if h preserves orienttion, ω h ω if h reverses orienttion. Proof. Write the one form s ω f 1 dx 1 + f 2 dx 2 + + f n dx n. Suppose first tht h is orienttion preserving. Using the definition of the pth integrl nd the chnge of vribles formul for the Riemnn integrl, b ( ω f j (t j (t dt d c d c h ( n j1 ( n j1 ( n j1 ω. f j ( ( h(τ j( h(τ dτ f j ( ( h(τ ( j h (τ dτ

9.2. PATH INTEGRALS 61 If h is orienttion reversing it will swp the order of the limits on the integrl introducing minus sign. The detils, long with finishing the proof for piecewise smooth pths is left to the reder s Exercise 9.2.1. Due to this proposition, if we hve set Γ R n which is the imge of simple piecewise smooth pth ( [,b], then if we somehow indicte the orienttion, tht is, which direction we trverse the curve, in other words where we strt nd where we finish, then we cn just write Γ ω, without mentioning the specific. Recll tht simple mens tht restricted to (,b is one-to-one, tht is, it is one-to-one except perhps t the endpoints. We cn lso often relx the simple pth condition little bit. Tht is, s long s : [,b] R n is one-to-one except t finitely mny points. Tht is, there re only finitely mny points p R n such tht 1 (p is more thn one point. See the exercises. The issue bout the injectivity problem is illustrted by the following exmple: Exmple 9.2.12: Suppose : [,2π] R 2 is given by (t : ( cos(t,sin(t nd : [,2π] R 2 is given by (t : ( cos(2t,sin(2t. Notice tht ( [,2π] ( [,2π], nd we trvel round the sme curve, the unit circle. But goes round the unit circle once in the counter clockwise direction, nd goes round the unit circle twice (in the sme direction. Then ydx + xdy ydx + xdy 2π 2π ( ( sin(t ( sin(t + cos(tcos(t dt 2π, ( ( sin(2t ( 2sin(2t + cos(t ( 2cos(t dt 4π. Furthermore, for simple closed pth, it does not even mtter where we strt the prmetriztion. See the exercises. Pths cn be cut up or conctented s follows. The proof is direct ppliction of the dditivity of the Riemnn integrl, nd is left s n exercise. Proposition 9.2.13. Let : [,c] R n be piecewise smooth pth. For some b (,c, define the piecewise smooth pths α [,b] nd [b,c]. For one-form ω defined on the curve defined by we hve ω ω + ω. α 9.2.3 Line integrl of function Sometimes we wish to simply integrte function ginst the so clled rc length mesure.

62 CHAPTER 9. ONE DIMENSIONAL INTEGRALS IN SEVERAL VARIABLES Definition 9.2.14. Suppose : [,b] R n is smooth pth, nd f is continuous function defined on the imge ( [,b]. Then define f ds : b f ( (t (t dt. The definition for piecewise smooth pth is similr s before nd is left to the reder. The geometric ide of this integrl is to find the re under the grph of function s we move round the pth. The line integrl of function is lso independent of the prmetriztion, but in this cse, the orienttion does not mtter. Proposition 9.2.15. Let : [,b] R n be piecewise smooth pth nd h: [c,d] R n smooth reprmetriztion. Suppose f is continuous function defined on the set ( [,b]. Then h f ds Proof. Suppose first tht h is orienttion preserving nd is smooth pth. Then s before f ds b d c d c d c h f (t (t dt f ds. f ( h(τ ( h(τ h (τdτ f ( h(τ ( h(τ h (τ dτ f ( h(τ ( h (τ dτ f ds. If h is orienttion reversing it will swp the order of the limits on the integrl but you lso hve to introduce minus sign in order to hve h inside the norm. The detils, long with finishing the proof for piecewise smooth pths is left to the reder s Exercise 9.2.2. Similrly s before, becuse of this proposition, if is simple, it does not mtter which prmetriztion we use. Therefore, if Γ ( [,b] we cn simply write Γ In this cse we lso do not need to worry bout orienttion, either wy we get the sme thing. f ds.

9.2. PATH INTEGRALS 63 Exmple 9.2.16: Let f (x,y x. Let C R 2 be hlf of the unit circle for x. We wish to compute f ds. C Prmetrize C by : [ π/2, π/2] R 2 given by (t ( cos(t,sin(t. Then (t ( sin(t,cos(t, nd π/2 ( sin(t 2 ( π/2 2 f ds f ds cos(t + cos(t dt cos(tdt 2. C π/2 π/2 Definition 9.2.17. Suppose Γ R n is prmetrized by simple piecewise smooth pth : [,b] R n, tht is ( [,b] Γ. The we define the length by l(γ : Γ ds ds b (t dt. Exmple 9.2.18: Let x,y R n be two points nd write [x,y] s the stright line segment between the two points x nd y. We cn prmetrize [x,y] by (t : (1 tx +ty for t running between nd 1. We note tht (t y x nd therefore we compute l ( [x,y] [x,y] ds 1 y x dt y x. So the length of [x,y] is the distnce between x nd y in the eucliden metric. A simple piecewise smooth pth : [,r] R n is sid to be n rc length prmetriztion if l ( ( [,t] t (τ dτ t. You cn think of such prmetriztion s moving round your curve t speed 1. 9.2.4 Exercises Exercise 9.2.1: Finish the proof of Proposition 9.2.11 for orienttion reversing reprmetriztions nd b piecewise smooth pths. Exercise 9.2.2: Finish the proof of Proposition 9.2.15 for orienttion reversing reprmetriztions nd b piecewise smooth pths. Exercise 9.2.3: Prove Proposition 9.2.13. Exercise 9.2.4: Suppose : [,b] R n is piecewise smooth pth, nd f is continuous function defined on the imge ( [,b]. Provide definition of f ds.

64 CHAPTER 9. ONE DIMENSIONAL INTEGRALS IN SEVERAL VARIABLES Exercise 9.2.5: Compute the length of the unit squre from Exmple 9.2.2 using the given prmetriztion. Exercise 9.2.6: Suppose : [,1] R n is piecewise smooth pth, nd ω is one-form defined on the imge ( [,b]. For r [,1], let r : [,r] R n be defined s simply the restriction of to [,r]. Show tht the function h(r : r ω is continuously differentible function on [,1]. Exercise 9.2.7: Suppose α : [,b] R n nd : [b,c] R n re piecewise smooth pths with α(b (b. Let : [,c] R n be defined by { α(t if t [,b], (t : (t if t (b,c]. Show tht is piecewise smooth pth, nd tht if ω is one-form defined on the curve given by, then ω ω + ω. α Exercise 9.2.8: Suppose : [,b] R n nd : [c,d] R n re two simple piecewise smooth closed pths. Tht is ( (b nd (c (d nd the restrictions (,b nd (c,d re one-to-one. Suppose Γ ( [,b] ( [c,d] nd ω is one-form defined on Γ R n. Show tht either ω ω, or ω ω. In prticulr, the nottion Γ ω mkes sense if we indicte the direction in which the integrl is evluted. Exercise 9.2.9: Suppose : [,b] R n nd : [c,d] R n re two piecewise smooth pths which re oneto-one except t finitely mny points. Tht is, there is t most finitely mny points p R n such tht 1 (p or 1 (p contins more thn one point. Suppose Γ ( [,b] ( [c,d] nd ω is one-form defined on Γ R n. Show tht either ω ω, or ω ω. In prticulr, the nottion Γ ω mkes sense if we indicte the direction in which the integrl is evluted.

9.3. PATH INDEPENDENCE 65 9.3 Pth independence Note:??? lectures 9.3.1 Pth independent integrls Let U R n be set nd ω one-form defined on U, nd x nd y two points in U. The integrl of ω from x to y is sid to be pth independent if for ny two piecewise smooth pths : [,b] R n nd : [c,d] R n such tht ( (c x nd (b (d y we hve In this cse we simply write y x ω ω. ω ω Not every one-form gives pth independent integrl. In fct, most do not. ω. Exmple 9.3.1: Let : [,1] R 2 be the pth (t (t, going from (, to (1,. : [,1] R 2 be the pth (t ( t,(1 tt lso going between the sme points. Then ydx ydx 1 1 1 2 (t 1(tdt 2 (t 1(tdt 1 (1dt, (1 tt(1dt 1 6. Let So (1, ydx is not pth independent. (, Definition 9.3.2. Let U R n be n open set nd f : U R continuously differentible function. Then the one-form d f : f dx 1 + f dx 2 + + f dx n x 1 x 2 x n is clled the totl derivtive of f. An open set U R n is sid to be pth connected if for every two points x nd y in U, there exists piecewise smooth pth strting t x nd ending t y. We will leve s n exercise tht every connected open set is pth connected. Proposition 9.3.3. Let U R n be pth connected open set nd ω one-form defined on U. Then y x ω

66 CHAPTER 9. ONE DIMENSIONAL INTEGRALS IN SEVERAL VARIABLES is pth independent for ll x,y U if nd only if there exists continuously differentible f : U R such tht ω d f. In fct, if such n f exists, then for ny two pints x,y U y ω f (y f (x. In other words if we fix x, then f (x C + x x ω. x Proof. First suppose tht the integrl is pth independent. Pick x U nd define f (x x x ω. Let e j be n rbitrry stndrd bsis vector. Compute f (x + he j f (x 1 ( x+he j ω h h x x x 1 h x+he j which follows by Proposition 9.2.13 nd pth indepdendence s x+he j x ω x x ω + x+he j x ω. Write ω ω 1 dx 1 +ω 2 dx 2 + +ω n dx n. Now pick the simplest pth possible from x to x+he j, tht is (t x +the j for t [,1]. Notice tht (t hs only simple nonzero component nd tht is the jth component which is h. Therefore 1 x+he j ω 1 1 1 ω j (x +the j hdt ω j (x +the j dt. h x h We wish to tke the limit s h. The function ω j is continuous. So given ε >, h cn be smll enough so tht ω(x ω j (x +the j < ε. Therefore for such smll h we find tht 1 ω j (x +the j dt ω(x < ε. Tht is f (x + he j f (x lim ω j (x, h h which is wht we wnted tht is d f ω. As ω j re continuous for ll j, we find tht f hs continuous prtil derivtives nd therefore is continuously differentible. For the other direction suppose f exists such tht d f ω. Suppose we tke smooth pth : [,b] U such tht ( x nd (b y, then b ( f ( d f (t x 1 (t + f ( (t 1 x 2 (t + + f ( (t 2 x n (t dt n b d [ ( ] f (t dt dt f (y f (x. x ω,

9.3. PATH INDEPENDENCE 67 Notice tht the vlue of the integrl only depends on x nd y, not the pth tken. Therefore the integrl is pth independent. We leve checking this for piecewise smooth pth s n exercise to the reder. This lso proves the lst ssertion since Then x x ω f (x f (x, letting C f (x finishes the proof. Proposition 9.3.4. Let U R n be pth connected open set nd ω 1-form defined on U. Then ω d f for some continuously differentible f : U : R if nd only if ω for every closed pth : [, b] U. Proof. Suppose first tht ω d f nd let be closed pth. Then we from bove we hve tht ω f ( (b f ( (, becuse ( (b for closed pth. Now suppose tht for every piecewise smooth closed pth, ω. Let x,y be two points in U nd let α : [,1] U nd : [,1] U be two piecewise smooth pths with α( ( x nd α(1 (1 y. Then let : [,2] U be defined by { α(t if t [,1], (t : (2 t if t (1,2]. This is piecewise smooth closed pth nd so ω α ω ω. This follows first by Proposition 9.2.13, nd then noticing tht the second prt is trvelled bckwrds so tht we get minus the integrl. Thus the integrl of ω on U is pth independent. There is locl criterion, tht is differentil eqution, tht gurntees pth independence. Tht is, under the right condition there exists n ntiderivtive f whose totl derivtive is the given one form ω. However, since the criterion is locl, we only get the result loclly. We cn define the ntiderivtive in ny so-clled simply connected domin, which informlly is domin with no 1-dimensionl holes to go round. To mke mtters simple, the usul wy this result is proved is for so-clled str-shped domins. Definition 9.3.5. Let U R n be n open set nd x U. We sy U is str shped domin with respect to x if for ny other point x U, the line segment between x nd x is in U, tht is, if (1 tx +tx U for ll t [,1]. If we sy simply str shped then U is str shped with respect to some x U.

68 CHAPTER 9. ONE DIMENSIONAL INTEGRALS IN SEVERAL VARIABLES Notice the difference between str shped nd convex. A convex domin is str shped, but str shped domin need not be convex. Theorem 9.3.6 (Poincrè lemm. Let U R n be str shped domin nd ω continuously differentible one-form defined on U. Tht is, if ω ω 1 dx 1 + ω 2 dx 2 + + ω n dx n, then ω 1,ω 2,...,ω n re continuously differentible functions. Suppose tht for every j nd k we hve ω j ω k, x k x j then there exists twice continuously differentible function f : U R such tht d f ω. The condition on the derivtives of ω is precisely the condition tht the second prtil derivtives commute. Tht is, if d f ω, then ω j 2 f. x k x k x j Proof. Suppose U is str shped with respect to y (y 1,y 2,...,y n U. Given x (x 1,x 2,...,x n U, define the pth : [,1] U s (t (1 ty +tx, so (t y x. Then let f (x ω 1 ( n ( ω k (1 ty +tx (yk x k k1 Now we cn differentite in x j under the integrl. We cn do tht since everything, including the prtils themselves re continuous. (( f 1 n ω k ( ( (x (1 ty +tx t(yk x k ω j (1 ty +tx dt x j k1 x j (( 1 n ω j ( ( (1 ty +tx t(yk x k ω j (1 ty +tx dt k1 x k 1 d [ ( ] tω j (1 ty +tx dt dt ω j (x. And this is precisely wht we wnted. Exmple 9.3.7: Without some hypothesis on U the theorem is not true. Let ω(x,y y x 2 + y 2 dx + x x 2 + y 2 dy dt

9.3. PATH INDEPENDENCE 69 be defined on R 2 \ {}. It is esy to see tht [ ] y y x 2 + y 2 [ x x x 2 + y 2 However, there is no f : R 2 \ {} R such tht d f ω. We sw in if we integrte from (1, to (1, long the unit circle, tht is (t ( cos(t,sin(t for t [,2π] we got 2π nd not s it should be if the integrl is pth independent or in other words if there would exist n f such tht d f ω. ]. 9.3.2 Vector fields A common object to integrte is so-clled vector field. Tht is n ssignment of vector t ech point of domin. Definition 9.3.8. Let U R n be set. A continuous function v: U R n is clled vector field. Write v (v 1,v 2,...,v n Given smooth pth : [,b] R n with ( [,b] U we define the pth integrl of the vectorfield v s b v d : v ( (t (tdt, where the dot in the definition is the stndrd dot product. Agin the definition of piecewise smooth pth is done by integrting over ech smooth intervl nd dding the result. If we unrvel the definition we find tht v d v 1 dx 1 + v 2 dx 2 + + v n dx n. Therefore wht we know bout integrtion of one-forms crries over to the integrtion of vector fields. For exmple pth independence for integrtion of vector fields is simply tht y x v d is pth independent (for ny if nd only if v f, tht is the grdient of function. The function f is then clled the potentil for v. A vector field v whose pth integrls re pth independent is clled conservtive vector field. The nming comes from the fct tht such vector fields rise in physicl systems where certin quntity, the energy is conserved.

7 CHAPTER 9. ONE DIMENSIONAL INTEGRALS IN SEVERAL VARIABLES 9.3.3 Exercises Exercise 9.3.1: Find n f : R 2 R such tht d f xe x2 +y 2 dx + ye x2 +y 2 dy. Exercise 9.3.2: Finish the proof of Proposition 9.3.3, tht is, we only proved the second direction for smooth pth, not piecewise smooth pth. Exercise 9.3.3: Show tht str shped domin U R n is pth connected. Exercise 9.3.4: Show tht U : R 2 \{(x,y R 2 : x,y } is str shped nd find ll points (x,y U such tht U is str shped with respect to (x,y. Exercise 9.3.5: Let : [,b] R n be simple nonclosed pth (so is one-to-one. Suppose tht ω is continuously differentible one-form defined on some open set V with ( [,b] V nd ω j x k ω k x j for ll j nd k. Prove tht there exists n open set U with ( [,b] U V nd twice continuously differentible function f : U R such tht d f ω. Hint 1: ( [,b] is compct. Hint 2: Piecing together severl different functions f cn be tricky, but notice tht the intersection of ny number of blls is lwys convex s blls re convex, nd convex sets re in prticulr connected (pth connected. Exercise 9.3.6: Show tht connected open set is pth connected. Hint: Strt with two points x nd y in connected set U, nd let U x U is the set of points tht re rechble by pth from x nd similrly for U y. Show tht both sets re open, since they re nonempty (x U x nd y U y it must be tht U x U y U. b Prove the converse tht is, pth connected set U R n is connected. Hint: for contrdiction ssume there exist two open nd disjoint nonempty open sets nd then ssume there is piecewise smooth (nd therefore continuous pth between point in one to point in the other. Exercise 9.3.7 (Hrd: Tke ω(x,y y x 2 + y 2 dx + x x 2 + y 2 dy defined on R 2 \ {(,}. Let : [,b] R 2 \ {(,} be closed piecewise smooth pth. Let R : {(x,y R 2 : x nd y }. Suppose tht R ( [,b] is finite set of k points. Then ω 2πl for some integer l with l k. Hint 1: First prove tht for pth tht strts nd end on R but does not intersect it otherwise, you find tht ω is 2π,, or 2π. Hint 2: You proved bove tht R2 \ R is str shped. Note: The number l is clled the winding number it it mesures how mny times does wind round the origin in the clockwise direction.

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