9.3 Path independence

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1 66 CHAPTER 9. ONE DIMENSIONAL INTEGRALS IN SEVERAL VARIABLES 9.3 Path independence Note:??? lectures Path independent integrals Let U R n be a set and ω a one-form defined on U, The integral of ω is said to be path independent if for any two points,y U and any two piecewise smooth paths : [a,b] U and : [c,d] U such that (a = (c = and (b = (d = y we have In this case we simply write y ω = ω = ω = Not every one-form gives a path independent integral. In fact, most do not. Eample 9.3.1: Let : [,1] R 2 be the path (t = (t, going from (, to (1,. : [,1] R 2 be the path (t = ( t,(1 tt also going between the same points. Then Let yd = yd = (t 1(tdt 2 (t 1(tdt 1 (1dt =, (1 tt(1dt = 1 6. So the integral of yd is not path independent. In particular, (1, yd does not make sense. (, Definition Let U R n be an open set and f : U R a continuously differentiable function. Then the one-form d f := f d 1 + f d f d n 1 2 n is called the total derivative of f. An open set U R n is said to be path connected if for every two points and y in U, there eists a piecewise smooth path starting at and ending at y. We will leave as an eercise that every connected open set is path connected. Normally only a continuous path is used in this definition, but for open sets two two definitions are equivalent. See the eercises.

2 9.3. PATH INDEPENDENCE 67 Proposition Let U R n be a path connected open set and ω a one-form defined on U. Then y ω is path independent (for all,y U if and only if there eists a continuously differentiable f : U R such that ω = d f. In fact, if such an f eists, then for any two points,y U y ω = f (y f (. In other words if we fi, then f ( = C + Proof. First suppose that the integral is path independent. Pick U and define f ( = Let e j be an arbitrary standard basis vector. Compute f ( + he j f ( = 1 ( +he j ω h h ω = 1 h +he j which follows by Proposition and path indepdendence as +he j ω = ω + +he j Write ω = ω 1 d 1 +ω 2 d 2 + +ω n d n. Now pick the simplest path possible from to +he j, that is (t = +the j for t [,1]. Notice that (t has only a simple nonzero component and that is the jth component which is h. Therefore +he j 1 1 ω = 1 ω j ( +the j hdt = ω j ( +the j dt. h h We wish to take the limit as h. The function ω j is continuous. So given ε >, h can be small enough so that ω( ω j ( +the j < ε. Therefore for such small h we find that 1 ω j ( +the j dt ω( < ε. That is f ( + he j f ( lim = ω j (, h h which is what we wanted that is d f = As ω j are continuous for all j, we find that f has continuous partial derivatives and therefore is continuously differentiable. For the other direction suppose f eists such that d f = Suppose we take a smooth path : [a,b] U such that (a = and (b = y, then b ( f ( d f = (t a 1 (t + f ( (t 1 2 (t + + f 2 b d [ ( ] = f (t dt a dt = f (y f (. 1 ω, ( (t n (t n dt

3 68 CHAPTER 9. ONE DIMENSIONAL INTEGRALS IN SEVERAL VARIABLES The value of the integral only depends on and y, not the path taken. Therefore the integral is path independent. We leave checking this for a piecewise smooth path as an eercise to the reader. Proposition Let U R n be a path connected open set and ω a 1-form defined on U. Then ω = d f for some continuously differentiable f : U : R if and only if ω = for every piecewise smooth closed path : [a,b] U. Proof. Suppose first that ω = d f and let be a piecewise smooth closed path. Then we from above we have that ω = f ( (b f ( (a =, because (a = (b for a closed path. Now suppose that for every piecewise smooth closed path, ω =. Let,y be two points in U and let α : [,1] U and : [,1] U be two piecewise smooth paths with α( = ( = and 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 a piecewise smooth closed path and so = ω = α ω This follows first by Proposition 9.2.1, and then noticing that the second part is travelled backwards so that we get minus the integral. Thus the integral of ω on U is path independent. There is a local criterion, that is a differential equation, that guarantees path independence. That is, under the right condition there eists an antiderivative f whose total derivative is the given one form However, since the criterion is local, we only get the result locally. We can define the antiderivative in any so-called simply connected domain, which informally is a domain where any path between two points can be continuously deformed into any other path between those two points. To make matters simple, the usual way this result is proved is for so-called star-shaped domains. Definition Let U R n be an open set and U. We say U is a star shaped domain with respect to if for any other point U, the line segment between and is in U, that is, if (1 t +t U for all t [,1]. If we say simply star shaped then U is star shaped with respect to some U.

4 9.3. PATH INDEPENDENCE 69 Notice the difference between star shaped and conve. A conve domain is star shaped, but a star shaped domain need not be conve. Theorem (Poincarè lemma. Let U R n be a star shaped domain and ω a continuously differentiable one-form defined on U. That is, if ω = ω 1 d 1 + ω 2 d ω n d n, then ω 1,ω 2,...,ω n are continuously differentiable functions. Suppose that for every j and k ω j k = ω k j, then there eists a twice continuously differentiable function f : U R such that d f = The condition on the derivatives of ω is precisely the condition that the second partial derivatives commute. That is, if d f = ω, then ω j = 2 f. k k j Proof. Suppose U is star shaped with respect to y = (y 1,y 2,...,y n U. Given = ( 1, 2,..., n U, define the path : [,1] U as (t = (1 ty +t, so (t = y. Then let f ( = ω = 1 ( n ( ω k (1 ty +t (yk k k=1 Now we can differentiate in j under the integral. We can do that since everything, including the partials themselves are continuous. (( f 1 n ω k ( ( ( = (1 ty +t t(yk k ω j (1 ty +t dt j k=1 j (( 1 n ω j ( ( = (1 ty +t t(yk k ω j (1 ty +t dt k=1 k 1 d [ ( ] = tω j (1 ty +t dt dt = ω j (. And this is precisely what we wanted. Eample 9.3.7: Without some hypothesis on U the theorem is not true. Let ω(,y = y 2 + y 2 d y 2 dy dt

5 7 CHAPTER 9. ONE DIMENSIONAL INTEGRALS IN SEVERAL VARIABLES be defined on R 2 \ {}. It is easy to see that [ ] y y 2 + y 2 = [ 2 + y 2 However, there is no f : R 2 \ {} R such that d f = We saw in if we integrate from (1, to (1, along the unit circle, that is (t = ( cos(t,sin(t for t [,2π] we got 2π and not as it should be if the integral is path independent or in other words if there would eist an f such that d f = ] Vector fields A common object to integrate is a so-called vector field. That is an assignment of a vector at each point of a domain. Definition Let U R n be a set. A continuous function v: U R n is called a vector field. Write v = (v 1,v 2,...,v n Given a smooth path : [a,b] R n with ( [a,b] U we define the path integral of the vectorfield v as b v d := v ( (t (tdt, a where the dot in the definition is the standard dot product. Again the definition of a piecewise smooth path is done by integrating over each smooth interval and adding the result. If we unravel the definition we find that v d = v 1 d 1 + v 2 d v n d n. Therefore what we know about integration of one-forms carries over to the integration of vector fields. For eample path independence for integration of vector fields is simply that y v d is path independent (so for any if and only if v = f, that is the gradient of a function. The function f is then called the potential for v. A vector field v whose path integrals are path independent is called a conservative vector field. The naming comes from the fact that such vector fields arise in physical systems where a certain quantity, the energy is conserved.

6 9.3. PATH INDEPENDENCE Eercises Eercise 9.3.1: Find an f : R 2 R such that d f = e 2 +y 2 d + ye 2 +y 2 dy. Eercise 9.3.2: Finish the proof of Proposition 9.3.3, that is, we only proved the second direction for a smooth path, not a piecewise smooth path. Eercise 9.3.3: Show that a star shaped domain U R n is path connected. Eercise 9.3.4: Show that U := R 2 \{(,y R 2 :,y = } is star shaped and find all points (,y U such that U is star shaped with respect to (,y. Eercise 9.3.5: Let : [a,b] R n be a simple nonclosed path (so is one-to-one. Suppose that ω is a continuously differentiable one-form defined on some open set V with ( [a,b] V and ω j k = ω k j for all j and k. Prove that there eists an open set U with ( [a,b] U V and a twice continuously differentiable function f : U R such that d f = Hint 1: ( [a,b] is compact. Hint 2: Piecing together several different functions f can be tricky, but notice that the intersection of any number of balls is always conve as balls are conve, and conve sets are in particular connected (path connected. Eercise 9.3.6: a Show that a connected open set is path connected. Hint: Start with two points and y in a connected set U, and let U U is the set of points that are reachable by a path from and similarly for U y. Show that both sets are open, since they are nonempty ( U and y U y it must be that U = U y = U. b Prove the converse that is, a path connected set U R n is connected. Hint: for contradiction assume there eist two open and disjoint nonempty open sets and then assume there is a piecewise smooth (and therefore continuous path between a point in one to a point in the other. Eercise 9.3.7: Usually path connectedness is defined using just continuous paths rather than piecewise smooth paths. Prove that the definitions are equivalent, in other words prove the following statement: Suppose U R n is such that for any,y U, there eists a continuous function : [a,b] U such that (a = and (b = y. Then U is path connected (in other words, then there eists a piecewise smooth path. Eercise (Hard: Take ω(,y = y 2 + y 2 d y 2 dy defined on R 2 \ {(,}. Let : [a,b] R 2 \ {(,} be a closed piecewise smooth path. Let R := {(,y R 2 : and y = }. Suppose that R ( [a,b] is a finite set of k points. Then ω = 2πl for some integer l with l k. Hint 1: First prove that for a path that starts and end on R but does not intersect it otherwise, you find that ω is 2π,, or 2π. Hint 2: You proved above that R2 \ R is star shaped. Note: The number l is called the winding number it it measures how many times does wind around the origin in the clockwise direction.

= 1 2 x (x 1) + 1 {x} (1 {x}). [t] dt = 1 x (x 1) + O (1), [t] dt = 1 2 x2 + O (x), (where the error is not now zero when x is an integer.

= 1 2 x (x 1) + 1 {x} (1 {x}). [t] dt = 1 x (x 1) + O (1), [t] dt = 1 2 x2 + O (x), (where the error is not now zero when x is an integer. Problem Sheet,. i) Draw the graphs for [] and {}. ii) Show that for α R, α+ α [t] dt = α and α+ α {t} dt =. Hint Split these integrals at the integer which must lie in any interval of length, such as [α,