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Textbook by: Lawrence Baggett. E-mail the author

# Vector Fields, Differential Forms, and Line Integrals

Module by: Lawrence Baggett. E-mail the author

Summary: We motivate our third definition of an integral over a curve by returning to physics. This definition is very much a real variable one, so that we think of the plane as R2R2 instead of C.C. A connection between this real variable definition and the complex variable definition of a contour integral will emerge later.

We motivate our third definition of an integral over a curve by returning to physics. This definition is very much a real variable one, so that we think of the plane as R2R2 instead of C.C. A connection between this real variable definition and the complex variable definition of a contour integral will emerge later.

Definition 1:

By a vector field on an open subset UU of R2,R2, we mean nothing more than a continuous function V(x,y)(P(x,y),Q(x,y))V(x,y)(P(x,y),Q(x,y)) from UU into R2.R2. The functions PP and QQ are called the components of the vector field V.V.

We will also speak of smooth vector fields, by which we will mean vector fields VV both of whose component functions PP and QQ have continuous partial derivatives

t i a l P t i a l x , t i a l P t i a l y , t i a l Q t i a l x a n d t i a l Q t i a l y t i a l P t i a l x , t i a l P t i a l y , t i a l Q t i a l x a n d t i a l Q t i a l y
(1)

on U.U.

The idea from physics is to think of a vector field as a force field, i.e., something that exerts a force at the point (x,y)(x,y) with magnitude |V(x,y)||V(x,y)| and acting in the direction of the vector V(x,y).V(x,y). For a particle to move within a force field, “work” must be done, that is energy must be provided to move the particle against the force, or energy is given to the particle as it moves under the influence of the force field. In either case, the basic definition of work is the product of force and distance traveled. More precisely, if a particle is moving in a direction uu within a force field, then the work done on the particle is the product of the component of the force field in the direction of uu and the distance traveled by the particle in that direction. That is, we must compute dot products of the vectors V(x,y)V(x,y) and u(x,y).u(x,y). Therefore, if a particle is moving along a curve C,C, parameterized with respect to arc length by γ:[0,L]C,γ:[0,L]C, and we write γ(t)=(x(t),y(t)),γ(t)=(x(t),y(t)), then the work W(z1,z2)W(z1,z2) done on the particle as it moves from z1=γ(0)z1=γ(0) to z2=γ(L)z2=γ(L) within the force field V,V, should intuitively be given by the formula

W ( z 1 , z 2 ) = 0 L V ( γ ( t ) ) γ ' ( t ) d t = 0 L P ( x ( t ) , y ( t ) ) x ' ( t ) + Q ( x ( t ) , y ( t ) ) y ' ( t ) d t C P d x + Q d y , W ( z 1 , z 2 ) = 0 L V ( γ ( t ) ) γ ' ( t ) d t = 0 L P ( x ( t ) , y ( t ) ) x ' ( t ) + Q ( x ( t ) , y ( t ) ) y ' ( t ) d t C P d x + Q d y ,
(2)

where the last expression is explicitly defining the shorthand notation we will be using.

The preceding discussion leads us to a new notion of what kind of object should be “integrated” over a curve.

Definition 2:

A differential form on a subset UU of R2R2 is denoted by ω=Pdx+Qdy,ω=Pdx+Qdy, and is determined by two continuous real-valued functions PP and QQ on U.U. We say that ωω is bounded or uniformly continuous if the functions PP and QQ are bounded or uniformly continuous functions on U.U. We say that the differential form ωω is smooth of order kk if the set UU is open, and the functions PP and QQ have continuous mixed partial derivatives of order k.k.

If ω=Pdx+Qdyω=Pdx+Qdy is a differential form on a set U,U, and if CC is any piecewise smooth curve of finite length contained in U,U, then we define the line integralCωCω of ωω over CC by

C ω = C P d x + Q d y = 0 L P ( γ ( t ) ) x ' ( t ) + Q ( γ ( t ) ) y ' ( t ) d t , C ω = C P d x + Q d y = 0 L P ( γ ( t ) ) x ' ( t ) + Q ( γ ( t ) ) y ' ( t ) d t ,
(3)

where γ(t)=(x(t),y(t))γ(t)=(x(t),y(t)) is a parameterization of CC by arc length.

REMARK There is no doubt that the integral in this definition exists, because PP and QQ are continuous functions on the compact set C,C, hence bounded, and γ'γ' is integrable, implying that both x'x' and y'y' are integrable. Therefore P(γ(t))x'(t)+Q(γ(t))y'(t)P(γ(t))x'(t)+Q(γ(t))y'(t) is integrable on (0,L).(0,L).

These differential forms ωω really should be called “differential 1-forms.” For instance, an example of a differential 2-form would look like Rdxdy,Rdxdy, and in higher dimensions, we could introduce notions of differential forms of higher and higher orders, e.g., in 3 dimension things like Pdxdy+Qdzdy+Rdxdz.Pdxdy+Qdzdy+Rdxdz. Because we will always be dealing with R2,R2, we will have no need for higher order differential forms, but the study of such things is wonderful. Take a course in Differential Geometry!

Again, we must see how this quantity CωCω depends, if it does, on different parameterizations. As usual, it does not.

## Exercise 1

Suppose ω=Pdx+Qdyω=Pdx+Qdy is a differential form on a subset UU of R2.R2.

1. Let CC be a piecewise smooth curve of finite length contained in UU that joins z1z1 to z2.z2. Prove that
Cω=CPdx+Qdy=abP(φ(t))x'(t)+Q(φ(t))y'(t)dtCω=CPdx+Qdy=abP(φ(t))x'(t)+Q(φ(t))y'(t)dt
(4)
for any parameterization φ:[a,b]Cφ:[a,b]C having components x(t)x(t) and y(t).y(t).
2. Let CC be as in part (a), and let C^C^ denote the reverse of C,C, i.e., the same set CC but thought of as a curve joining z2z2 to z1.z1. Show that c^ω=-Cω.c^ω=-Cω.
3. Let CC be as in part (a). Prove that
|CPdx+Qdy|(MP+MQ)L,|CPdx+Qdy|(MP+MQ)L,
(5)
where MPMP and MQMQ are bounds for the continuous functions |P||P| and |Q||Q| on the compact set C,C, and where LL is the length of C.C.

## Example 1

The simplest interesting example of a differential form is constructed as follows. Suppose UU is an open subset of R2,R2, and let f:URf:UR be a differentiable real-valued function of two real variables; i.e., both of its partial derivatives exist at every point (x,y)U.(x,y)U. (See the last section of Chapter IV.) Define a differential form ω=df,ω=df, called the differential of f,f, by

d f = t i a l f t i a l x d x + t i a l f t i a l y d y , d f = t i a l f t i a l x d x + t i a l f t i a l y d y ,
(6)

i.e., P=tialf/tialxP=tialf/tialx and Q=tialf/tialy.Q=tialf/tialy. These differential forms dfdf are called exact differential forms.

REMARK Not every differential form ωω is exact, i.e., of the form df.df. Indeed, determining which ωω's are dfdf's boils down to what may be the simplest possible partial differential equation problem. If ωω is given by two functions PP and Q,Q, then saying that ω=dfω=df amounts to saying that ff is a solution of the pair of simultaneous partial differential equations

t i a l f t i a l x = P and t i a l f t i a l y = Q . t i a l f t i a l x = P and t i a l f t i a l y = Q .
(7)

See part (b) of the exercise below for an example of a nonexact differential form.

Of course if a real-valued function ff has continuous partial derivatives of the second order, then (Reference) tells us that the mixed partials fxyfxy and fyxfyx must be equal. So, if ω=Pdx+Qdy=dfω=Pdx+Qdy=df for some such f,f, Then PP and QQ would have to satisfy tialP/tialy=tialQ/tialx.tialP/tialy=tialQ/tialx. Certainly not every PP and QQ would satisfy this equation, so it is in fact trivial to find examples of differential forms that are not differentials of functions. A good bit more subtle is the question of whether every differential form Pdx+Qdy,Pdx+Qdy, for which tialP/tialy=tialQ/tialx,tialP/tialy=tialQ/tialx, is equal to some df.df. Even this is not true in general, as part (c) of the exercise below shows. The open subset UU on which the differential form is defined plays a significant role, and, in fact, differential forms provide a way of studying topologically different kinds of open sets.

In fact, although it may seem as if a differential form is really nothing more than a pair of functions, the concept of a differential form is in part a way of organizing our thoughts about partial differential equation problems into an abstract mathematical context. This abstraction is a good bit more enlightening in higher dimensional spaces, i.e., in connection with functions of more than two variables. Take a course in Multivariable Analysis!

## Exercise 2

1. Solve the pair of simultaneous partial differential equations
tialftialx=x+yandtialftialy=x-y.tialftialx=x+yandtialftialy=x-y.
(8)
2. Show that it is impossible to solve the pair of simultaneous partial differential equations
tialftialx=x+yandtialftialy=y3.tialftialx=x+yandtialftialy=y3.
(9)
Hence, conclude that the differential form ω=(x+y)dx+y3dyω=(x+y)dx+y3dy is not the differential dfdf of any real-valued function f.f.
3. Let UU be the open subset of R2R2 that is the complement of the single point (0,0).(0,0). Let P(x,y)=-y/(x2+y2)P(x,y)=-y/(x2+y2) and Q(x,y)=x/(x2+y2).Q(x,y)=x/(x2+y2). Show that tialP/tialy=tialQ/tialxtialP/tialy=tialQ/tialx at every point of U,U, but that ω=Pdx+Qdyω=Pdx+Qdy is not the differential dfdf of any smooth function ff on U.U. HINT: If PP were fx,fx, then ff would have to be of the form f(x,y)=-tan-1(x/y)+g(y),f(x,y)=-tan-1(x/y)+g(y), where gg is some differentiable function of y.y. Show that if Q=fyQ=fy then g(y)g(y) is a constant c.c. Hence, f(x,y)f(x,y) must be -tan-1(x/y)+c.-tan-1(x/y)+c. But this function ff is not continuous, let alone differentiable, at the point (1,0).(1,0). Consider limf(1,1/n)limf(1,1/n) and limf(1,-1/n).limf(1,-1/n).

The next thing we wish to investigate is the continuity of CωCω as a function of the curve C.C. This brings out a significant difference in the concepts of line integrals versis integrals with respect to arc length. For the latter, we typically think of a fixed curve and varying functions, whereas with line integrals, we typically think of a fixed differential form and variable curves. This is not universally true, but should be kept in mind.

## Theorem 1

Let ω=Pdx+Qdyω=Pdx+Qdy be a fixed, bounded, uniformly continuous differential form on a set UU in R2,R2, and let CC be a fixed piecewise smooth curve of finite length LL, parameterized by φ:[a,b]C,φ:[a,b]C, that is contained in U.U. Then, given an ϵ>0ϵ>0 there exists a δ>0δ>0 such that, for any curve C^C^ contained in U,U,|Cω-C^ω|<ϵ|Cω-C^ω|<ϵ whenever the following conditions on the curve C^C^ hold:

1.  C^C^ is a piecewise smooth curve of finite length L^L^ contained in U,U, parameterized by φ^:[a,b]C^.φ^:[a,b]C^.
2.  |φ(t)-φ^(t)|<δ|φ(t)-φ^(t)|<δ for all t[a,b].t[a,b].
3.  ab|φ'(t)-φ^'(t)|dt<δ.ab|φ'(t)-φ^'(t)|dt<δ.

### Proof

Let ϵ>0ϵ>0 be given. Because both PP and QQ are bounded on U,U, let MPMP and MQMQ be upper bounds for the functions |P||P| and |Q||Q| respectively. Also, since both PP and QQ are uniformly continuous on U,U, there exists a δ>0δ>0 such that if |(c,d)-(c',d')|<δ,|(c,d)-(c',d')|<δ, then |P(c,d)-P(c',d')|<ϵ/4L|P(c,d)-P(c',d')|<ϵ/4L and |Q(c,d)-Q(c',d')|<ϵ/4L.|Q(c,d)-Q(c',d')|<ϵ/4L. We may also choose this δδ to be less than both ϵ/4MPϵ/4MP and ϵ/4MQ.ϵ/4MQ. Now, suppose C^C^ is a curve of finite length L^,L^, parameterized by φ^:[a,b]C^,φ^:[a,b]C^, and that |φ(t)-φ^(t)|<δ|φ(t)-φ^(t)|<δ for all t[a,b],t[a,b], and that ab|φ'(t)-φ^'(t)|<δ.ab|φ'(t)-φ^'(t)|<δ. Writing φ(t)=(x(t),y(t))φ(t)=(x(t),y(t)) and φ^(t)=(x^(t),y^(t)),φ^(t)=(x^(t),y^(t)), we have

0 | C P d x + Q d y - C ^ P d x + Q d y | = | a b P ( φ ( t ) ) x ' ( t ) - P ( φ ^ ( t ) ) x ^ ' ( t ) + Q ( φ ( t ) ) y ' ( t ) - Q ( φ ^ ( t ) ) y ^ ' ( t ) d t | a b | P ( φ ( t ) ) x ' ( t ) - P ( φ ^ ( t ) ) x ^ ' ( t ) | d t + a b | Q ( φ ( t ) ) y ' ( t ) - Q ( φ ^ ( t ) ) y ^ ' ( t ) | d t a b | P ( φ ( t ) ) - P ( φ ^ ( t ) ) | | x ' ( t ) | d t + a b | P ( φ ^ ( t ) ) | | x ' ( t ) - x ^ ' ( t ) | d t + a b | Q ( φ ( t ) ) - Q ( φ ^ ( t ) ) | | y ' ( t ) | d t + a b | Q ( φ ^ ( t ) ) | | y ' ( t ) - y ^ ' ( t ) | d t ϵ 4 L a b | x ' ( t ) | d t + M P a b | x ' ( t ) - x ^ ' ( t ) | d t + ϵ 4 L a b | y ' ( t ) | d t + M Q a b | y ' ( t ) - y ^ ' ( t ) | d t ϵ 4 L a b | φ ' ( t ) | d t + M P a b | φ ' ( t ) - φ ^ ' ( t ) | d t + ϵ 4 L a b | φ ' ( t ) | d t + M Q a b | φ ' ( t ) - φ ^ ' ( t ) | d t < ϵ 4 + ϵ 4 + M P δ + M Q δ < ϵ , 0 | C P d x + Q d y - C ^ P d x + Q d y | = | a b P ( φ ( t ) ) x ' ( t ) - P ( φ ^ ( t ) ) x ^ ' ( t ) + Q ( φ ( t ) ) y ' ( t ) - Q ( φ ^ ( t ) ) y ^ ' ( t ) d t | a b | P ( φ ( t ) ) x ' ( t ) - P ( φ ^ ( t ) ) x ^ ' ( t ) | d t + a b | Q ( φ ( t ) ) y ' ( t ) - Q ( φ ^ ( t ) ) y ^ ' ( t ) | d t a b | P ( φ ( t ) ) - P ( φ ^ ( t ) ) | | x ' ( t ) | d t + a b | P ( φ ^ ( t ) ) | | x ' ( t ) - x ^ ' ( t ) | d t + a b | Q ( φ ( t ) ) - Q ( φ ^ ( t ) ) | | y ' ( t ) | d t + a b | Q ( φ ^ ( t ) ) | | y ' ( t ) - y ^ ' ( t ) | d t ϵ 4 L a b | x ' ( t ) | d t + M P a b | x ' ( t ) - x ^ ' ( t ) | d t + ϵ 4 L a b | y ' ( t ) | d t + M Q a b | y ' ( t ) - y ^ ' ( t ) | d t ϵ 4 L a b | φ ' ( t ) | d t + M P a b | φ ' ( t ) - φ ^ ' ( t ) | d t + ϵ 4 L a b | φ ' ( t ) | d t + M Q a b | φ ' ( t ) - φ ^ ' ( t ) | d t < ϵ 4 + ϵ 4 + M P δ + M Q δ < ϵ ,
(10)

as desired.

Again, we have a special notation when the curve CC is a graph. If g:[a,b]Rg:[a,b]R is a piecewise smooth function, then its graph CC is a piecewise smooth curve, and we write graph(g)Pdx+Qdygraph(g)Pdx+Qdy for the line integral of the differential form Pdx+QdyPdx+Qdy over the curve C=graph(g).C=graph(g).

As alluded to earlier, there is a connection between contour integrals and line integrals. It is that a single contour integral can often be expressed in terms of two line integrals. Here is the precise statement.

## Theorem 2

Suppose CC is a piecewise curve of finite length, and that f=u+ivf=u+iv is a complex-valued, continuous function on C.C. Let φ:[a,b]Cφ:[a,b]C be a parameterization of C,C, and write φ(t)=x(t)+iy(t).φ(t)=x(t)+iy(t). Then

C f ( ζ ) d ζ = C ( U d x - v d y ) + C ( v d x + u d y ) . C f ( ζ ) d ζ = C ( U d x - v d y ) + C ( v d x + u d y ) .
(11)

### Proof

We just compute:

C f ( ζ ) d ζ = a b f ( φ ( t ) ) φ ' ( t ) d t = a b ( u ( φ ( t ) ) + i v ( φ ( t ) ) ) ( x ' ( t ) + i y ' ( t ) ) d t = a b ( u ( φ ( t ) ) x ' ( t ) - v ( φ ( t ) ) y ' ( t ) ) + i ( v ( φ ( t ) ) x ' ( t ) + u ( φ ( t ) ) y ' ( t ) ) d t = a b ( u ( φ ( t ) ) x ' ( t ) - v ( φ ( t ) ) y ' ( t ) ) d t + i a b ( v ( φ ( t ) ) x ' ( t ) + u ( φ ( t ) ) y ' ( t ) ) d t = C u d x - v d y + i C v d x + u d y , C f ( ζ ) d ζ = a b f ( φ ( t ) ) φ ' ( t ) d t = a b ( u ( φ ( t ) ) + i v ( φ ( t ) ) ) ( x ' ( t ) + i y ' ( t ) ) d t = a b ( u ( φ ( t ) ) x ' ( t ) - v ( φ ( t ) ) y ' ( t ) ) + i ( v ( φ ( t ) ) x ' ( t ) + u ( φ ( t ) ) y ' ( t ) ) d t = a b ( u ( φ ( t ) ) x ' ( t ) - v ( φ ( t ) ) y ' ( t ) ) d t + i a b ( v ( φ ( t ) ) x ' ( t ) + u ( φ ( t ) ) y ' ( t ) ) d t = C u d x - v d y + i C v d x + u d y ,
(12)

as asserted.

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