Notes on lecture 17
1 Identifying conservative vector fields
For discussing this topic, we first need to understand the concepts of an
open region, a connected region, and a simply-connected region. Let us first
understand the concepts of an open region and a connected region. We will
study simply-connected regions in the next lecture.
Definition 1.0.1. A region D in the plane is called open if for every point
P in D, there exists a disk with center at P that lies entirely in D. The
region D is called connected if any two points in D can be joined by a path
that lies entirely in D. ⇤
Theorem 1.0.2. Suppose ~
R F is a vector field that is continuous on an open
~ is independent of path in D, then F~ is a
connected region D. If C F~ · dr
conservative vector field on D, that is, there exists a function f such that
Of = F~ .
Proof: Let (a, b) be a fixed point in D. We construct the desired potential
function f by defining
Z (x,y)
f (x, y) = ~
F~ · dr
(a,b)
R
for any point (x, y) in D. Since C F~ · dr~ is independent of path, it does not
matter which path C from (a, b) to (x, y) is used to evaluate f (x, y).
Since D is open, there exists a disk contained in D with center (x, y). Choose
any point (x1 , y) in the disk with x1 < x and let C consist of any path C1
from (a, b) to (x1 , y) followed by the horizontal line segment C2 from (x1 , y)
to (x, y). Please see Figure 1.0.1 for this path. Such a path exists because D
is a connected region.
Then
1
, Figure 1.0.1: The curve C from (a, b) to (x, y)
Z Z Z (x1 ,y) Z
f (x, y) = ~ +
F~ · dr ~ =
F~ · dr ~ +
F~ · dr ~
F~ · dr.
C1 C2 (a,b) C2
Notice that the first of these integrals does not depend on x, so
Z
@ @ ~
f (x, y) = 0 + F~ · dr.
@x @x C2
If we write F~ = P î + Qĵ, then
Z Z
~ =
F~ · dr P dx + Qdy.
C2 C2
On C2 , y is constant, so dy = 0. Using t as a parameter, where x1 t x,
we have
Z Z x
@ @ @
f (x, y) = P dx + Qdy = P (t, y)dt = P (x, y),
@x @x C2 @x x1
where the last equality follows from the fact that
Z x
@
P ( )d = P (x).
@x a
A similar argument, using a vertical line segment (see Figure 1.0.2) shows
that
@
f (x, y) = Q(x, y).
@y
2
, Figure 1.0.2: Another curve for the proof of Theorem 1.0.2
Thus
@f @f
F~ = P î + Qĵ = î + ĵ = Of,
@x @y
which says that F~ is conservative. ⇤
Question: How is it possible to determine whether or not a vector field F~
is conservative?
Suppose it is known that F~ = P î + Qĵ is conservative, where P and Q
have continuous first order partial derivatives. Then there exists a function
f such that F~ = Of , that is,
@f @f
P = and Q = .
@x @y
Therefore, by Clairaut’s theorem,
@P @ 2f @ 2f @Q
= = = .
@y @y@x @x@y @x
This proves the following theorem:
Theorem 1.0.3. If F~ (x, y) = P (x, y)î + Q(x, y)ĵ is a conservative vector
field, where P and Q have continuous first order partial derivatives on a
domain D, then throughout D we have
@P @Q
= .
@y @x
3
1 Identifying conservative vector fields
For discussing this topic, we first need to understand the concepts of an
open region, a connected region, and a simply-connected region. Let us first
understand the concepts of an open region and a connected region. We will
study simply-connected regions in the next lecture.
Definition 1.0.1. A region D in the plane is called open if for every point
P in D, there exists a disk with center at P that lies entirely in D. The
region D is called connected if any two points in D can be joined by a path
that lies entirely in D. ⇤
Theorem 1.0.2. Suppose ~
R F is a vector field that is continuous on an open
~ is independent of path in D, then F~ is a
connected region D. If C F~ · dr
conservative vector field on D, that is, there exists a function f such that
Of = F~ .
Proof: Let (a, b) be a fixed point in D. We construct the desired potential
function f by defining
Z (x,y)
f (x, y) = ~
F~ · dr
(a,b)
R
for any point (x, y) in D. Since C F~ · dr~ is independent of path, it does not
matter which path C from (a, b) to (x, y) is used to evaluate f (x, y).
Since D is open, there exists a disk contained in D with center (x, y). Choose
any point (x1 , y) in the disk with x1 < x and let C consist of any path C1
from (a, b) to (x1 , y) followed by the horizontal line segment C2 from (x1 , y)
to (x, y). Please see Figure 1.0.1 for this path. Such a path exists because D
is a connected region.
Then
1
, Figure 1.0.1: The curve C from (a, b) to (x, y)
Z Z Z (x1 ,y) Z
f (x, y) = ~ +
F~ · dr ~ =
F~ · dr ~ +
F~ · dr ~
F~ · dr.
C1 C2 (a,b) C2
Notice that the first of these integrals does not depend on x, so
Z
@ @ ~
f (x, y) = 0 + F~ · dr.
@x @x C2
If we write F~ = P î + Qĵ, then
Z Z
~ =
F~ · dr P dx + Qdy.
C2 C2
On C2 , y is constant, so dy = 0. Using t as a parameter, where x1 t x,
we have
Z Z x
@ @ @
f (x, y) = P dx + Qdy = P (t, y)dt = P (x, y),
@x @x C2 @x x1
where the last equality follows from the fact that
Z x
@
P ( )d = P (x).
@x a
A similar argument, using a vertical line segment (see Figure 1.0.2) shows
that
@
f (x, y) = Q(x, y).
@y
2
, Figure 1.0.2: Another curve for the proof of Theorem 1.0.2
Thus
@f @f
F~ = P î + Qĵ = î + ĵ = Of,
@x @y
which says that F~ is conservative. ⇤
Question: How is it possible to determine whether or not a vector field F~
is conservative?
Suppose it is known that F~ = P î + Qĵ is conservative, where P and Q
have continuous first order partial derivatives. Then there exists a function
f such that F~ = Of , that is,
@f @f
P = and Q = .
@x @y
Therefore, by Clairaut’s theorem,
@P @ 2f @ 2f @Q
= = = .
@y @y@x @x@y @x
This proves the following theorem:
Theorem 1.0.3. If F~ (x, y) = P (x, y)î + Q(x, y)ĵ is a conservative vector
field, where P and Q have continuous first order partial derivatives on a
domain D, then throughout D we have
@P @Q
= .
@y @x
3
