Using Divergence and Curl

Learning Objectives

Use the properties of curl and divergence to determine whether a vector field is conservative.

Now that we understand the basic concepts of divergence and curl, we can discuss their properties and establish relationships between them and conservative vector fields.

F

${\bf{F}}$ is a vector field in

R^{3}

$\mathbb{R}^3$ , then the curl of

F

${\bf{F}}$ is also a vector field in

R^{3}

$\mathbb{R}^3$ . Therefore, we can take the divergence of a curl. The next theorem says that the result is always zero. This result is useful because it gives us a way to show that some vector fields are not the curl of any other field. To give this result a physical interpretation, recall that divergence of a velocity field

v

${\bf{v}}$ at point

P

$P$ measures the tendency of the corresponding fluid to flow out of

P

$P$ . Since

div curl (v) = 0

$\text{div curl }({\bf{v}})=0$ , the net rate of flow in vector field curl(

v

${\bf{v}}$ ) at any point is zero. Taking the curl of vector field

F

${\bf{F}}$ eliminates whatever divergence was present in

F

${\bf{F}}$ .

Theorem: divergence of the curl

Let ${\bf{F}}=\langle{P},Q,R\rangle$ be a vector field in $\mathbb{R}^3$ such that the component functions all have continuous second-order partial derivatives. Then, $\text{div curl }({\bf{F}})=\nabla\cdot(\nabla\times{\bf{F}})=0$ .

Proof

By the definitions of divergence and curl, and by Clairaut’s theorem,

$\large{\begin{aligned} \text{div curl }({\bf{F}})&=\text{div }[(R_y-Q_z){\bf{i}}+(P_z-R_x){\bf{j}}+(Q_x-P_y){\bf{k}}] \\&=R_{yx}-Q_{xz}+P_{yz}-R_{yx}+Q_{zx}-P_{zy} \\ &=0 \end{aligned}}$ .

$_\blacksquare$

Example: showing that a vector field is not the curl of another

Show that ${\bf{F}}(x,y,z)=e^x{\bf{i}}+yz{\bf{j}}+xz^2{\bf{k}}$ is not the curl of another vector field. That is, show that there is no other vector ${\bf{G}}$ with curl ${\bf{G}}={\bf{F}}$ .

Show Solution

Notice that the domain of ${\bf{F}}$ is all of $\mathbb{R}^3$ and the second-order partials of ${\bf{F}}$ are all continuous. Therefore, we can apply the previous theorem to ${\bf{F}}$ .

The divergence of ${\bf{F}}$ is $e^x+z+2xz$ . If ${\bf{F}}$ were the curl of vector field $\bf{G}$ , then $\text{div }{\bf{F}}=\text{div curl }{\bf{G}}=0$ . But, the divergence of ${\bf{F}}$ is not zero, and therefore ${\bf{F}}$ is not the curl of any other vector field.

try it

Is it possible for ${\bf{G}}(x,y,z)=\langle\sin{x},\cos{y},\sin{(x,y,z)}\rangle$ to be the curl of a vector field?

Show Solution

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

You can view the transcript for “CP 6.45” here (opens in new window).

With the next two theorems, we show that if ${\bf{F}}$ is a conservative vector field then its curl is zero, and if the domain of ${\bf{F}}$ is simply connected then the converse is also true. This gives us another way to test whether a vector field is conservative.

theorem: curl of a conservative vector field

If ${\bf{F}}=\langle{P},Q,R\rangle$ is conservative, then $\text{curl } \bf{F}=\bf{0}$ $curl F = 0 .$

Proof

Since conservative vector fields satisfy the cross-partials property, all the cross-partials of ${\bf{F}}$ are equal. Therefore,

$\large{\begin{aligned} \text{curl }{\bf{F}}&=(R_y-Q_z){\bf{i}}+(P_z-R_x){\bf{j}}+(Q_x-P_y){\bf{k}} \\ &=0 \end{aligned}}$ .

$_\blacksquare$

The same theorem is true for vector fields in a plane.

Since a conservative vector field is the gradient of a scalar function, the previous theorem says that $\text{curl }(\nabla{f})=0$ for any scalar function $f$ . In terms of our curl notation, $\nabla\times\nabla{(f)}=0$ . This equation makes sense because the cross product of a vector with itself is always the zero vector. Sometimes equation $\nabla\times\nabla{(f)}=0$ is simplified as $\nabla\times\nabla=0$ .

theorem: curl test for a conservative field

Let ${\bf{F}}=\langle{P},Q,R\rangle$ be a vector field in space on a simply connected domain. If $\text{curl }{\bf{F}}=0$ , then ${\bf{F}}$ is conservative.

Proof

Since $\text{curl }{\bf{F}}=0$ , we have that $R_y=Q_z$ , $P_z=R_x$ , and $Q_x=P_y$ . Therefore, ${\bf{F}}$ satisfies the cross-partials property on a simply connected domain, and Cross-Partial Property of Conservative Fields Theorem implies that ${\bf{F}}$ is conservative.

$_\blacksquare$

The same theorem is also true in a plane. Therefore, if ${\bf{F}}$ is a vector field in a plane or in space and the domain is simply connected, then ${\bf{F}}$ is conservative if and only if $\text{curl }{\bf{F}}=0$ .

Example: testing whether a vector field is conservative

Use the curl to determine whether ${\bf{F}}(x,y,z)=\langle{y}z,xz,xy\rangle$ is conservative.

Show Solution

Note that the domain of ${\bf{F}}$ is all of $\mathbb{R}^3$ , which is simply connected (Figure 1). Therefore, we can test whether ${\bf{F}}$ is conservative by calculating its curl.

<img src="/apps/archive/20220422.171947/resources/7749bed210536f0229f8267b69c34794d0305459" data-media-type="image/jpeg" alt="A diagram showing the curl of a vector field in two dimensions. The curl is zero. The arrows seem to be pointing up and over into the yz plane." id="30">

Figure 1. The curl of vector field ${\bf{F}}(x,y,z)=\langle{y}z,xz,xy\rangle$ is zero.

The curl of ${\bf{F}}$ is

$\left(\frac{\partial}{\partial{y}}xy-\frac{\partial}{\partial{z}}xz\right){\bf{i}}+\left(\frac{\partial}{\partial{y}}yz-\frac{\partial}{\partial{z}}xy\right){\bf{j}}+\left(\frac{\partial}{\partial{y}}xz-\frac{\partial}{\partial{z}}yz\right){\bf{k}}=(x-x){\bf{i}}+(y-y){\bf{j}}+(z-z){\bf{k}}=0$ .

Thus, ${\bf{F}}$ is conservative.

We have seen that the curl of a gradient is zero. What is the divergence of a gradient? If $f$ is a function of two variables, then $\text{div }(\nabla{f})=\nabla\cdot(\nabla{f})=f_{xx}+f_{yy}$ . We abbreviate this “double dot product” as $\nabla^2$ . This operator is called the , and in this notation Laplace’s equation becomes $\nabla^2f=0$ . Therefore, a harmonic function is a function that becomes zero after taking the divergence of a gradient.

Similarly, if $f$ is a function of three variables then

$\large{\text{div }(\nabla{f})=\nabla\cdot(\nabla{f})=f_{xx}+f_{yy}+f_{zz}}$ .

Using this notation we get Laplace’s equation for harmonic functions of three variables:

$\large{\nabla^2f=0}$ .

Harmonic functions arise in many applications. For example, the potential function of an electrostatic field in a region of space that has no static charge is harmonic.

Example: analyzing a function

Is it possible for $f(x, y)=x^{2}+x-y$ to be the potential function of an electrostatic field that is located in a region of $\mathbb{R}^2$ free of static charge?

Show Solution

If $f$ were such a potential function, then $f$ would be harmonic. Note that $f_{xx}=2$ , and $f_{yy}=0$ , so $f_{xx}+f_{yy}\ne0$ . Therefore, $f$ is not harmonic and $f$ cannot represent an electrostatic potential.

try it

Is it possible for function $f(x, y)=x^{2}-y^{2}+x$ to be the potential function of an electrostatic field located in a region of $\mathbb{R}^2$ free of static charge?

Show Solution

Module 6: Vector Calculus