Kenneth Kuttler

30.5. CONSERVATIVE VECTOR FIELDS 605

30.5 Conservative Vector FieldsDefinition 30.5.1 A vector field F defined in a three dimensional region is said tobe conservative2 if for every piecewise smooth closed curve C, it follows

∫C F ·dR= 0.

This looks a little different than the earlier definition. However, the main result in thissection is an assertion that these are exactly the same.

Definition 30.5.2 Let (x,p1, · · · ,pn,y) be an ordered list of points in Rp. Letp(x,p1, · · · ,pn,y) denote the piecewise smooth curve consisting of a straight line segmentfrom x to p1 and then the straight line segment from p1 to p2 · · · and finally the straightline segment from pn to y. This is called a polygonal curve. An open set in Rp, U, is said tobe a region if it has the property that for any two points x,y ∈U, there exists a polygonalcurve joining the two points.

Conservative vector fields are important because of the following theorem, sometimescalled the fundamental theorem for line integrals.

Theorem 30.5.3 Let U be a region in Rp and let F : U → Rp be a continuousvector field. Then F is conservative if and only if there exists a scalar valued function of pvariables φ such that F = ∇φ . Furthermore, if C is an oriented curve which goes from xto y in U, then ∫

CF · dR= φ (y)−φ (x) . (30.3)

Thus the line integral is path independent in this case. This function φ is called a scalarpotential for F .

Proof: To save space and fussing over minutia, denote by p(x0,x) a polygonal curvefrom x0 to x. Thus the orientation is such that itgoes from x0 to x. The curve p(x,x0) denotes the same set of points but in the oppositeorder. Suppose first F is conservative. Fix x0 ∈U and let

φ (x)≡∫p(x0,x)

F ·dR.

This is well defined because if q (x0,x) is another polygonal curve joining x0 to x, Thenthe curve obtained by following p(x0,x) from x0 to x and then from x to x0 alongq (x,x0) is a closed piecewise smooth curve and so by assumption, the line integral alongthis closed curve equals 0. However, this integral is just∫

p(x0,x)F ·d R+

∫q(x,x0)

F ·d R=∫p(x0,x)

F ·d R−∫q(x0,x)

F ·dR

which shows ∫p(x0,x)

F ·d R=∫q(x0,x)

F ·dR

and that φ is well defined. For small t,

φ (x + tei)−φ (x)

∫p(x0,x+tei)

F ·d R−∫p(x0,x)

F ·dRt

∫p(x0,x)

F ·d R+∫p(x,x+tei)

F ·d R−∫p(x0,x)

F ·dRt

2There is no such thing as a liberal vector field.

30.5. CONSERVATIVE VECTOR FIELDS 60530.5 Conservative Vector FieldsDefinition 30.5.1 4 vector field F defined in a three dimensional region is said tobe conservative? if for every piecewise smooth closed curve C, it follows Jc F-dR=0.This looks a little different than the earlier definition. However, the main result in thissection is an assertion that these are exactly the same.Definition 30.5.2 Le: (@,P1,°**,Pn,Y) be an ordered list of points in R?. Letp(X,P1,°** ;Pn,y) denote the piecewise smooth curve consisting of a straight line segmentfrom & to p, and then the straight line segment from p, to p--: and finally the straightline segment from p,, to y. This is called a polygonal curve. An open set in R”, U, is said tobe a region if it has the property that for any two points x,y € U, there exists a polygonalcurve joining the two points.Conservative vector fields are important because of the following theorem, sometimescalled the fundamental theorem for line integrals.Theorem 30.5.3 Let U be a region in R? and let F : U — R? be a continuousvector field. Then F is conservative if and only if there exists a scalar valued function of pvariables @ such that F = Vo. Furthermore, if C is an oriented curve which goes from &to y in U, then[F-4R=0(y)-9(@). (30.3)Thus the line integral is path independent in this case. This function @ is called a scalarpotential for F.Proof: To save space and fussing over minutia, denote by p(ao,x) a polygonal curvefrom a to x. Thus the orientation is such that itgoes from ao to x. The curve p(x, 2%) denotes the same set of points but in the oppositeorder. Suppose first F’ is conservative. Fix xo € U and let6 (a) =|. Pak.This is well defined because if g(a, a) is another polygonal curve joining ao to x, Thenthe curve obtained by following p(ao,x) from xp to x and then from x to xo alongq(x,Zq) is a closed piecewise smooth curve and so by assumption, the line integral alongthis closed curve equals 0. However, this integral is just| F-dR+ | F-dR= | F-dR- | F-dRP(xo0,x) q(x,x0) P(x0,x) q(xo0,x)which shows| F-dR= F-dRp(xo,x) q(xo0,x)and that @ is well defined. For small r,g(a +te;)—o(@) _— Iplaowtre)) FI R= Ip(ay.a) F dRt ~ tSplao.n) FI R+ Sp(a,04te;) FIR In(ao.x) FIRt?There is no such thing as a liberal vector field.