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Section 2.3 When is a one-form exact?

Since not all one-forms can be written as differentials of functions, i.e. not all one-forms are exact, a natural question arises: can we determine, looking at a one-form, whether it is exact or not? Similarly, can we easily determine whether a vector field is conservative? Unfortunately we will not be able to answer this question at the moment, we will come back to it later. For the time being, we will be able to find a necessary condition for a one-form to be exact, which in the context of vector calculus is sometimes called a "screening test" for conservative vector fields.

Subsection 2.3.1 Closed one-forms in \(\mathbb{R}^2\)

Let us focus first on one-forms and vector fields on \(U \subset \mathbb{R}^2\text{.}\)

Definition 2.3.1. Closed one-forms in \(\mathbb{R}^2\).

We say that a one-form \(\omega = f\ dx + g\ dy\) on \(U \subset \mathbb{R}^2\) is closed if

\begin{equation*} \frac{\partial f}{\partial y} = \frac{\partial g}{\partial x}. \end{equation*}

This definition may seem ad hoc, but it is important because of the following lemma.

Suppose that \(\omega\) is exact: then it can be written as

\begin{equation*} \omega = f\ dx + g \ dy = d F = \frac{\partial F}{\partial x}\ dx + \frac{\partial F}{\partial y}\ dy \end{equation*}

for some smooth function \(F\) on \(U \subset \mathbb{R}^2\text{.}\) It will then be closed if

\begin{equation*} \frac{\partial f}{\partial y} = \frac{\partial^2 F}{\partial y \partial x} \end{equation*}

is equal to

\begin{equation*} \frac{\partial g}{\partial x} = \frac{\partial^2 F}{\partial x \partial y}. \end{equation*}

Equality of the two expressions follows from the Clairaut-Schwarz theorem, which states that partial derivatives commute, as long as they are continuous. But continuity of the partial derivatives is guaranteed by the fact that all partial derivatives of \(F\) exist and are continuous, since by definition (see Definition 2.1.1) one-forms are assumed to have smooth coefficients. Therefore the one-form is closed.

Note however that the conserve statement is not necessarily true: not all closed one-forms on \(U \subset \mathbb{R}^2\) are exact. In fact, the question of when closed one-forms are exact is an important one; the result is known as Poincare's lemma. We will come back to this in Section 3.7.

There is of course an analogous statement for conservative vector fields. The only minor difference is that we need to impose a condition on the component functions of vector fields, since vector fields are not always smooth by definition (see Definition Definition 2.1.2.

Same as for Lemma 2.3.2, but for the associated vector fields.

In the context of vector fields, it is called a screening test, because it is a quick test to determine whether a vector field has a chance at all to be conservative. In other words, if a vector field does not pass the screening test, then it is certainly not conservative. However, if it passes the screening test, at this stage we cannot conclude anything. Just like closed one-forms are not necessarily exact.

Consider the exact one-form on \(\mathbb{R}^2\) from Example 2.2.7, \(\omega = \cos x\ dx - \sin y\ dy\text{.}\) We show that it is closed, according to Definition 2.3.1. The partial derivatives are easily calculated:

\begin{equation*} \frac{\partial}{\partial y} \cos x = 0, \qquad \frac{\partial}{\partial x} (- \sin y) = 0. \end{equation*}

Thus Definition 2.3.1 is satisfied, and \(\omega\) is closed.

If a vector field is conservative, then it passes the screening test. Correspondingly, if a one-form is exact, then it is closed. But the converse statement is not necessarily true (we will revisit it later in Section 3.7). Consider for instance the one-form

\begin{equation*} \omega = - \frac{y}{x^2+y^2}\ dx + \frac{x}{x^2+y^2}\ dy. \end{equation*}

Calculating the partial derivatives for Definition 2.3.1, we get that

\begin{equation*} \frac{\partial}{\partial y}\left( - \frac{y}{x^2+y^2} \right) = \frac{y^2- x^2}{x^2+y^2}, \qquad \frac{\partial}{\partial x}\left( \frac{x}{x^2+y^2} \right) = \frac{ y^2-x^2}{x^2+y^2}. \end{equation*}

The two expressions are equal, and thus \(\omega\) is closed. However, one can show that \(\omega\) is not exact: there does not exist a function \(f\) such that \(\omega= d f\text{.}\)

Subsection 2.3.2 Closed one-forms in \(\mathbb{R}^3\)

We focused on \(\mathbb{R}^2\) for simplicity, but similar results hold for one-forms and vector fields in \(\mathbb{R}^3\text{.}\)

Definition 2.3.6. Closed one-forms in \(\mathbb{R}^3\).

We say that a one-form \(\omega = f\ dx + g\ dy + h\ dz\) on \(U \subset \mathbb{R}^3\) is closed if

\begin{equation*} \frac{\partial f}{\partial y} = \frac{\partial g}{\partial x}, \qquad \frac{\partial f}{\partial z} = \frac{\partial h}{\partial x}, \qquad \frac{\partial g}{\partial z} = \frac{\partial h}{\partial y}. \end{equation*}

We then have the lemma:

Suppose that \(\omega\) is exact: then it can be written as

\begin{equation*} \omega = f\ dx + g \ dy + h\ dz = d F = \frac{\partial F}{\partial x}\ dx + \frac{\partial F}{\partial y}\ dy + \frac{\partial F}{\partial z}\ dz \end{equation*}

for some smooth function \(F\) on \(U \subset \mathbb{R}^3\text{.}\) It will then be closed if

\begin{equation*} \frac{\partial^2 F}{\partial y \partial x} = \frac{\partial^2 F}{\partial x \partial y}, \qquad \frac{\partial^2 F}{\partial z \partial x} = \frac{\partial^2 F}{\partial x \partial z}, \qquad \frac{\partial^2 F}{\partial z \partial y} = \frac{\partial^2 F}{\partial y \partial z}. \end{equation*}

As before, these equalities follow from the Clairaut-Schwarz theorem, which states that partial derivatives commute as long as they are continuous.

The analogous statement for vector fields goes as follows:

Same as for Lemma 2.3.7, but for the associated vector fields.

Remark 2.3.9.

At this stage the definition of closeness for one-forms and the associated screening test for vector fields appear to be rather ad hoc. Sure, they are necessary conditions for a one-form to be exact and a vector field to be conservative, but is that it? No, not really. In fact, those conditions will come out very naturally when we go beyond one-forms and introduce the theory of \(n\)-forms in general. We will then see how we can in fact "differentiate" forms - this is the notion of exterior derivative. Then the definition of closed one-forms in Definition 2.3.1 will be simply that a one-form \(\omega\) is closed if its exterior derivative \(d \omega\) vanishes. Furthermore, if \(F\) is the vector field associated to a one-form \(\omega\text{,}\) then the vector field associated to its derivative \(d \omega\) will be called the "curl" of \(\mathbf{F}\text{,}\) and denoted by \(\boldsymbol{\nabla} \times \mathbf{F}\text{.}\) The screening test for vector fields will then be that \(\mathbf{F}\) is "curl-free", that is \(\boldsymbol{\nabla} \times \mathbf{F} = 0\text{.}\) All fun stuff, but it will have to wait for a bit! Coming in Chapter 4.

Exercises 2.3.3 Exercises

1.

Show that the one-form \(\omega = \cos(y)\ dx - x \sin(y)\ dy\) is closed. Is it exact?

Solution.

Let us write \(\omega = f\ dx + g\ dy\) for \(f(x,y) = \cos(y)\) and \(g(x,y) = - x \sin(y)\text{.}\) To show that it is closed, we calculate:

\begin{equation*} \frac{\partial f}{\partial y} = - \sin(y), \qquad \frac{\partial g}{\partial x} = - \sin(y). \end{equation*}

As the two partial derivatives are equal, we conclude that \(\omega\) is a closed one-form.

Is it exact? We are looking for a function \(F(x,y)\) such that \(d F = \omega\text{,}\) that is,

\begin{equation*} \frac{\partial F}{\partial x} = \cos(y), \qquad \frac{\partial F}{\partial y} = - x \sin(y). \end{equation*}

Integrating the first equation, we get \(F(x,y) = x \cos(y) + h(y)\) for some function \(h(y)\text{.}\) Substituting in the second equation, we get

\begin{equation*} - x \sin(y) + h'(y) = - x \sin(y), \end{equation*}

from which we conclude that \(h'(y) = 0\text{,}\) that is \(h(y) = C\text{.}\) We pick \(C=0\text{.}\) We conclude that \(\omega = d F\) with \(F(x,y) = x \cos(y)\text{,}\) and hence it is exact.

2.

Show that the vector field \(\mathbf{F}(x,y,z) = \left( y+z, x+z, x+y \right)\) passes the screening test.

Solution.

We simply need to calculate partial derivatives. If we denote the component functions by \((f_1,f_2,f_3)\text{,}\) we get:

\begin{gather*} \frac{\partial f_1}{\partial y} = 1, \qquad \frac{\partial f_1}{\partial z} = 1, \qquad \frac{\partial f_2}{\partial z} = 1\\ \frac{\partial f_2}{\partial x} = 1, \qquad \frac{\partial f_3}{\partial x} = 1, \qquad \frac{\partial f_3}{\partial y} = 1. \end{gather*}

It follows that \(\mathbf{F}\) passes the screening test. In fact, it is easy to show that it is conservative, with a potential given by \(f(x,y,z) = x y + x z + y z.\)

3.

Determine whether the one-form \(\omega = x\ dx + x\ dy + z\ dz\) is exact. If it is, find a function \(f\) such that \(\omega = df\text{.}\)

Solution.

If we write \(\omega = f\ dx + g\ dy + h\ dz\text{,}\) we see right away that

\begin{equation*} \frac{\partial f}{\partial y} = 0, \end{equation*}

while

\begin{equation*} \frac{\partial g}{\partial x} = 1. \end{equation*}

Thus \(\omega\) is not closed, and hence it cannot exact.

4.

Consider the vector field \(\mathbf{F}(x,y) = (x y^n , 2 x^2 y^3)\) for some positive integer \(n\text{.}\) Find the value of \(n\) for which \(\mathbf{F}\) is conservative on \(\mathbb{R}^2\text{,}\) and find a potential for this value of \(n\text{.}\)

Solution.

For \(\mathbf{F}\) to be conservative, it must pass the screening test. We write \(\mathbf{F} = (f_1, f_2)\text{,}\) and calculate:

\begin{equation*} \frac{\partial f_1}{\partial y} = n x y^{n-1}, \qquad \frac{\partial f_2}{\partial x} = 4 x y^3. \end{equation*}

We see that the two partial derivatives are equal for all \((x,y)\) if and only if \(n=4\text{,}\) in which case the vector field becomes \(\mathbf{F} = (x y^4, 2 x^2 y^3)\text{.}\) To show that it is conservative and find a potential function, we are looking for a function \(f(x,y)\) such that \(\boldsymbol{\nabla} f= (x y^4, 2 x^2 y^3).\) So we must have

\begin{equation*} \frac{\partial f}{\partial x}= x y^4, \qquad \frac{\partial f}{\partial y} = 2 x^2 y^3. \end{equation*}

Integrating the first equation, we get

\begin{equation*} f(x,y) = \frac{1}{2} x^2 y^4 + g(y) \end{equation*}

for some function \(g(y)\text{.}\) Substituting in the second equation, we get

\begin{equation*} 2 x^2 y^3 + g'(y) =2 x^2 y^3\text{,} \end{equation*}

from which we get \(g'(y) = 0\text{,}\) that is \(g(y) = C\text{.}\) We pick \(C=0\text{.}\) Thus \(\mathbf{F}\) is conservative for \(n=4\text{,}\) and a potential function is given by

\begin{equation*} f(x,y) = \frac{1}{2} x^2 y^4. \end{equation*}