Closed Vector Fields in the Plane

Section 8.3 Closed Vector Fields in the Plane

f = (f_{1}, f_{2})

is a vector field on a subset of the plane condition then (8.4) reduces to

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\begin{matrix} (8.5) & \frac{\partial}{\partial x_{2}} f_{1} (x) = \frac{\partial}{\partial x_{1}} f_{2} (x) \end{matrix}

for all

x

in the domain of

f .

We now give an example of a closed but not conservative vector field in the plane. The example will occur many times in the sequel, so make sure you understand it well.

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Example 8.12. A closed but not conservative vector field.

Let

f

be the vector field given by

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f (x, y) = (\frac{- y}{x^{2} + y^{2}}, \frac{x}{x^{2} + y^{2}})

for

(x, y) \neq (0, 0) .

It is shown in Figure 8.13.

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Figure 8.13. The vector field

f (x, y) := (\frac{- y}{x^{2} + y^{2}}, \frac{x}{x^{2} + y^{2}})

R^{2} .

Note that

‖ f ‖ = ‖ (x, y) ‖^{- 1},

so the length of the vectors go to infinity as

(x, y)

approaches

(0, 0) .

We say that

f

has a singularity at

(0, 0) .

We check whether

f

is closed

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Solution.

We have that

\begin{aligned} \frac{\partial}{\partial y} \frac{- y}{x^{2} + y^{2}} & = \frac{- (x^{2} + y^{2}) + 2 y^{2}}{(x^{2} + y^{2})^{2}} = \frac{y^{2} - x^{2}}{(x^{2} + y^{2})^{2}} \\ \frac{\partial}{\partial x} \frac{x}{x^{2} + y^{2}} & = \frac{x^{2} + y^{2} - 2 y^{2}}{(x^{2} + y^{2})^{2}} = \frac{y^{2} - x^{2}}{(x^{2} + y^{2})^{2}} . \end{aligned}

Hence

f

is indeed a closed vector field on

R^{2} ∖ {0} .

We now want to compute the integral along the circle of radius

R

centred at

(0, 0) .

Assume that the circle is oriented counterclockwise. We parametrise it by

\begin{aligned} γ (t) & = (R \cos t, R \sin t), & t \in [0, 2 π] . \end{aligned}

Then

γ^{'} (t) = (- R \sin t, R \cos t),

and using the definition of the line integral we see that

\begin{aligned} \int_{C} f \cdot d x & = \int_{0}^{2 π} \frac{R \sin t}{R^{2} \sin^{2} + R^{2} \cos^{t}} R \sin t + \int_{0}^{2 π} \frac{R \cos t}{R^{2} \sin^{2} + R^{2} \cos^{t}} R \cos t d t \\ = \int_{0}^{2 π} \frac{R^{2} \cos t + R^{2} \sin^{2}}{R^{2} \sin^{2} + R^{2} \cos^{t}} d t = \int_{0}^{2 π} 1 d t = 2 π . \end{aligned}

The circle is a closed curve. Hence if

f

were conservative the integral would be zero. As it is not zero,

f

cannot be conservative, even though it is closed. Note that

R^{2} ∖ {0}

is not simply connected, so Theorem 8.11 does not apply!

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