Gradient Fields

In this section, we study a special kind of vector field called a gradient field or a conservative field. These vector fields are extremely important in physics because they can be used to model physical systems in which energy is conserved. Gravitational fields and electric fields associated with a static charge are examples of gradient fields.

Recall that if $f$ is a (scalar) function of $x$ and $y$, then the gradient of $f$ is

$\large{\text{grad}f=\nabla{f}=f_x(x,y){\bf{i}}+f_y(x,y){\bf{j}}}$.

We can see from the form in which the gradient is written that $\nabla{f}$. is a vector field in $\mathbb{R}^2$. Similarly, if $f$ is a function of $x$, $y$, and $z$, then the gradient of $f$ is

$\large{\text{grad}f=\nabla{f}=f_x(x,y,z){\bf{i}}+f_y(x,y,z){\bf{j}}+f_z(x,y,z){\bf{k}}}$.

The gradient of a three-variable function is a vector field in $\mathbb{R}^3$.

A gradient field is a vector field that can be written as the gradient of a function, and we have the following definition.

Example: sketching a gradient vector field

Use technology to plot the gradient vector field of $f(x, y)=x^{2}y^{2}$.

Show Solution

The gradient of $f$ is $\nabla{f}=\langle2xy^2,2x^2y\rangle$. To sketch the vector field, use a computer algebra system such as Mathematica. Figure 1 shows $\nabla{f}$.

Figure 1. The gradient vector field is $\nabla{f}$, where $f(x, y)=x^{2}y^{2}$.

try it

Use technology to plot the gradient vector field of $f(x,y)=\sin{x}\cos{y}$.

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Figure 2. A virtual representation of the gradient vector field $f(x,y)=\sin{x}\cos{y}$.

Consider the function $f(x, y)=x^{2}y^{2}$ from Example “Sketching a Gradient Vector Field”. Figure 4 shows the level curves of this function overlaid on the function’s gradient vector field. The gradient vectors are perpendicular to the level curves, and the magnitudes of the vectors get larger as the level curves get closer together, because closely grouped level curves indicate the graph is steep, and the magnitude of the gradient vector is the largest value of the directional derivative. Therefore, you can see the local steepness of a graph by investigating the corresponding function’s gradient field.

Figure 3. The gradient field of $f(x, y)=x^{2}y^{2}$ and several level curves of $f$. Notice that as the level curves get closer together, the magnitude of the gradient vectors increases.

As we learned earlier, a vector field ${\bf{F}}$ is a conservative vector field, or a gradient field if there exists a scalar function $f$ such that $\nabla{f}={\bf{F}}$. In this situation, $f$ is called a potential function for ${\bf{F}}$. Conservative vector fields arise in many applications, particularly in physics. The reason such fields are called conservative is that they model forces of physical systems in which energy is conserved. We study conservative vector fields in more detail later in this chapter.

You might notice that, in some applications, a potential function $f$ for ${\bf{F}}$ is defined instead as a function such that $-\nabla{f}={\bf{F}}$. This is the case for certain contexts in physics, for example.

Example: verifying a potential function

The velocity of a fluid is modeled by field ${\bf{v}}(x,y)=\langle{x}y,\frac{x^2}2-y\rangle$. Verify that $f(x,y)=\frac{x^2y}2-\frac{y^2}2$ is a potential function for $\bf{v}$.

Show Solution

To show that $f$ is a potential function, we must show that $\nabla{f}=\bf{v}$. Note that $f_x=xy$ and $f_y=\frac{x^2}2-y$. Therefore, $\nabla{f}=\langle{x}y,\frac{x^2}2-y\rangle$ and $f$ is a potential function for ${\bf{v}}$ (Figure 4).

Figure 4. Velocity field ${\bf{v}}(x,y)$ has a potential function and is a conservative field.

Watch the following video to see the worked solution to the above Try It

If $\bf{F}$ is a conservative vector field, then there is at least one potential function $f$ such that $\nabla{f}=\bf{F}$. But, could there be more than one potential function? If so, is there any relationship between two potential functions for the same vector field? Before answering these questions, let’s recall some facts from single-variable calculus to guide our intuition. Recall that if $k(x)$ is an integrable function, then $k$ has infinitely many antiderivatives. Furthermore, if $F$ and $G$ are both antiderivatives of $k$, then $F$ and $G$ differ only by a constant. That is, there is some number $C$ such that $F(x)=G(x)+C$.

Now let $\bf{F}$ be a conservative vector field and let $f$ and $g$ be potential functions for $\bf{F}$. Since the gradient is like a derivative, $\bf{F}$ being conservative means that $\bf{F}$ is “integrable” with “antiderivatives” $f$ and $g$. Therefore, if the analogy with single-variable calculus is valid, we expect there is some constant $C$ such that $f(x)=g(x)+C$. The next theorem says that this is indeed the case.

To state the next theorem with precision, we need to assume the domain of the vector field is connected and open. To be connected means if $P_1$ and $P_2$ are any two points in the domain, then you can walk from $P_1$ to $P_2$ along a path that stays entirely inside the domain.

Proof

Since $f$ and $g$ are both potential functions for $\bf{F}$, then $\nabla{(f-g)}=\nabla{f}-\nabla{g}={\bf{F}}-{\bf{F}}=0$. Let $h=f-g$ then we have $\nabla{h}=0$. We would like to show that $h$ is a constant function.

Assume $h$ is a function of $x$ and $y$ (the logic of this proof extends to any number of independent variables). Since $\nabla{h}=0$, we have $h_x=0$ and $h_y=0$. The expression $h_x=0$ implies that $h$ is a constant function with respect to $x$—that is, $h(x, y)=k_1(y)$ for some function $k_1$.  Similarly, $h_y=0$ implies $h(x, y)=k_2(x)$ for some function $k_2$. Therefore, function $h$ depends only on $y$ and also depends only on $x$. Thus, $h(x, y)=C$ for some constant $C$ on the connected domain of $\bf{F}$. Note that we really do need connectedness at this point; if the domain of $\bf{F}$ came in two separate pieces, then $k$ could be a constant $C_1$ on one piece but could be a different constant $C_2$ on the other piece. Since $f-g=h=C$, we have that $f=g+C$, as desired.

$_\blacksquare$

Conservative vector fields also have a special property called the cross-partial property. This property helps test whether a given vector field is conservative.

Proof

Since $\bf{F}$ is conservative, there is a function $f(x, y)$ such that $\nabla{f}=\bf{F}$. Therefore, by the definition of the gradient, $f_x=P$ and $f_y=Q$. By Clairaut’s theorem, $f_{xy}=f_{yx}$, but, $f_{xy}=P_y$ and $f_{yx}=Q_x$, and thus $P_y=Q_x$.

$_\blacksquare$

Clairaut’s theorem gives a fast proof of the cross-partial property of conservative vector fields in $\mathbb{R}^3$, just as it did for vector fields in $\mathbb{R}^2$.

The Cross-Partial Property of Conservative Vector Fields Theorem shows that most vector fields are not conservative. The cross-partial property is difficult to satisfy in general, so most vector fields won’t have equal cross-partials.

We conclude this section with a word of warning: The Cross-Partial Property of Conservative Vector Fields Theorem says that if $\bf{F}$ is conservative, then $\bf{F}$ has the cross-partial property. The theorem does not say that, if $\bf{F}$ has the cross-partial property, then $\bf{F}$ is conservative (the converse of an implication is not logically equivalent to the original implication). In other words, The Cross-Partial Property of Conservative Vector Fields Theorem can only help determine that a field is not conservative; it does not let you conclude that a vector field is conservative. For example, consider vector field ${\bf{F}}(x,y)=\langle{x}^2y,\frac{x^3}3\rangle$. This field has the cross-partial property, so it is natural to try to use The Cross-Partial Property of Conservative Vector Fields Theorem to conclude this vector field is conservative. However, this is a misapplication of the theorem. We learn later how to conclude that $\bf{F}$ is conservative.