29 conservative fields potential functions

Conservative Fields, Potential Functions
and Path Independence

In the following discussion, we need open domains
that are simply connected, i.e. one piece (connected),
and don't have any hole (simple).

Not simpleNot connected Simply connected

Recalling a theorem about mixed partial derivatives:
"Given a real-valued function P(x, y) where the
partial derivatives Px, Py, Pxy and Pyx are continuous
in a simply connected D, then Pxy = Pyx in D."

Recalling a theorem about mixed partial derivatives:
"Given a real-valued function P(x, y) where the
partial derivatives Px, Py, Pxy and Pyx are continuous
in a simply connected D, then Pxy = Pyx in D."
We will call such a function a "nice" function.

Given a nice function P(x, y), its gradients F= P(x, y)
is a vector field.

is a vector field. (Recall that P(x, y) = fxi + fyj).

is a vector field. (Recall that P(x, y) = fxi + fyj).
P(x, y) is said to be a potential function of the field F.

is a vector field. (Recall that P(x, y) = Pxi + Pyj).
A vector field F that is the gradient field of a "nice"
function P(x, y) is called a conservative field.

Given an arbitrary vector field F, we like to know if it
is conservative, that is, if there is a potential function
P whose gradient is F.

Theorem: Given a vector field F = f(x, y)i + g(x, y)j
with f and g having continuous first partial derivatives
(in a open simply connected region) is conservative
if and only if fy = gx.
Given an arbitrary vector field F, we like to know if it
is conservative, that is, if there is a potential function
P whose gradient is F.

If a function P(x, y) is a "nice" function, its gradient
field P(x, y) = Pxi + Pyj satisfy Pxy = Pyx since the
mixed partials are the same.

This theorem gives the converse of the above fact.

If a vector field F = fi + gj satisfy fy = gx and they are
continuous, then F is the gradient field of a "nice"
function P(x, y).

function P(x, y).
Example: a. Show the vector field
F(x, y) = xy2
i + (y + yx2
)j is conservative.

function P(x, y).
F(x, y) = xy2
i + (y + yx2
)j is conservative.
f(x, y) = xy2
, g(x, y) = y + yx2

function P(x, y).
F(x, y) = xy2
i + (y + yx2
)j is conservative.
f(x, y) = xy2
, g(x, y) = y + yx2
fy = 2xy, gx = 2xy  fy = gx.

function P(x, y).
F(x, y) = xy2
i + (y + yx2
)j is conservative.
f(x, y) = xy2
, g(x, y) = y + yx2
fy = 2xy, gx = 2xy  fy = gx.
Hence F(x, y) is conservative.

b. Find a potential function P(x, y) such that
P(x, y) = Pxi + Pyj = F(x, y) = xy2
i + (y + yx2
)j.

i + (y + yx2
)j.
We recover the potential P(x, y) by partial integration.

i + (y + yx2
)j.
Since the gradient of P is to be F,
so Px = f(x, y) = xy2
.

i + (y + yx2
)j.
.
Therefore P = ∫ f(x, y) dx = ∫ xy2
dx

i + (y + yx2
)j.
.
dx
treating y as a constant, we've
P = x2
y2
/2 + C(y)
where C(y) is a function in y and is to be determined.

i + (y + yx2
)j.
.
But Py = g(x, y) = y + yx2
dx
P = x2
y2
/2 + C(y)

i + (y + yx2
)j.
.
= x2
y + Cy(y)
dx
P = x2
y2
/2 + C(y)

i + (y + yx2
)j.
.
= x2
y + Cy(y)  Cy(y) = y
dx
P = x2
y2
/2 + C(y)

i + (y + yx2
)j.
.
= x2
y + Cy(y)  Cy(y) = y
Hence C(y) = ∫ydy = y2
/2 + K
dx
P = x2
y2
/2 + C(y)

i + (y + yx2
)j.
.
= x2
y + Cy(y)  Cy(y) = y
Hence C(y) = ∫ydy = y2
/2 + K
dx
P = x2
y2
/2 + C(y)
So P(x, y) = x2
y2
/2 + y2
/2 + K

Fundemental Theorem of Line Integral

Given a conservative field F in a open simply
connected domain D and let P(x, y) be a potential
function of F .

function of F . Let (x0, y0) and (x1, y1) be two points in
D and C be any continuous curve from (x0, y0) to
(x1, y1).

(x1, y1).
(x0, y0)
(x1, y1)C

(x1, y1). Then the line integral
∫C
F • dC = P(x1, y1) – P(x0, y0)
(x0, y0)
(x1, y1)C

(x1, y1). Then the line integral
∫C
F • dC = P(x1, y1) – P(x0, y0)
(x0, y0)
(x1, y1)C1
C2
C3
From the theorem, the line
integrals in the figure
∫C1
F • dC = ∫C2
F • dC = ∫C3
F • dC
are the same in a conservative field F.

c. Use the fact that P(x, y) = x2
y2
/2 + y2
/2 + k is the
potential function of F(x, y) = xy2
i + (y + yx2
)j.
Find where
C = <cos(t), sin(t)> with
π/2 < t < π .
∫C
F • dC

y2
/2 + y2
/2 + k is the
i + (y + yx2
)j.
Find where
π/2 < t < π .
∫C
F • dC
C (0, 1)
(-1, 0)

y2
/2 + y2
/2 + k is the
i + (y + yx2
)j.
Find where
π/2 < t < π .
∫C
F • dC
Since F is conservative with potential P and
C(π/2) = (0, 1), C(π) = (-1, 0) as the starting and
ending points in the domain,
C (0, 1)
(-1, 0)

y2
/2 + y2
/2 + k is the
i + (y + yx2
)j.
Find where
π/2 < t < π .
∫C
F • dC
ending points in the domain, therefore
∫C
F • dC = P(-1, 0) – P(0, 1)
C (0, 1)
(-1, 0)

y2
/2 + y2
/2 + k is the
i + (y + yx2
)j.
Find where
π/2 < t < π .
∫C
F • dC
∫C
F • dC = P(-1, 0) – P(0, 1) = 0 – ½
C (0, 1)
(-1, 0)

y2
/2 + y2
/2 + k is the
i + (y + yx2
)j.
Find where
π/2 < t < π .
∫C
F • dC
∫C
F • dC = P(-1, 0) – P(0, 1) = 0 – ½ = -1/2.
C (0, 1)
(-1, 0)

Corrolary: Given a conservative field F over an open
simply connected region D and C is a closed loop in
D, then
∫C
F • dC = 0

D, then
∫C
F • dC = 0
C
D

D, then
∫C
F • dC = 0
(x0, y0)
=(x1, y1)
Proof: A closed loop is just a curve with the starting
point the same as the ending point.
C
D

D, then
∫C
F • dC = 0
(x0, y0)
=(x1, y1)
F is conservative so it has a
potential function P(x, y) and
∫C
F • dC = P(x0, y0) – P(x1 – y1)
C
D

D, then
∫C
F • dC = 0
(x0, y0)
=(x1, y1)
F is conservative so it has a
potential function P(x, y) and
∫C
F • dC = P(x0, y0) – P(x1 – y1)
= 0.
as shown.
C
D

29 conservative fields potential functions

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