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.

If [latex]{\bf{F}}[/latex] is a vector field in [latex]\mathbb{R}^3[/latex], then the curl of [latex]{\bf{F}}[/latex] is also a vector field in [latex]\mathbb{R}^3[/latex]. 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 [latex]{\bf{v}}[/latex] at point [latex]P[/latex] measures the tendency of the corresponding fluid to flow out of [latex]P[/latex]. Since [latex]\text{div curl }({\bf{v}})=0[/latex], the net rate of flow in vector field curl([latex]{\bf{v}}[/latex]) at any point is zero. Taking the curl of vector field [latex]{\bf{F}}[/latex] eliminates whatever divergence was present in [latex]{\bf{F}}[/latex].

Theorem: divergence of the curl

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

Proof

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

[latex]\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}}[/latex].

[latex]_\blacksquare[/latex]

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

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

try it

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

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

With the next two theorems, we show that if [latex]{\bf{F}}[/latex] is a conservative vector field then its curl is zero, and if the domain of [latex]{\bf{F}}[/latex] 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 [latex]{\bf{F}}=\langle{P},Q,R\rangle[/latex] is conservative, then [latex]\text{curl } \bf{F}=\bf{0}[/latex]. 

Proof

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

[latex]\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}}[/latex].

[latex]_\blacksquare[/latex]

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 [latex]\text{curl }(\nabla{f})=0[/latex] for any scalar function [latex]f[/latex]. In terms of our curl notation, [latex]\nabla\times\nabla{(f)}=0[/latex]. This equation makes sense because the cross product of a vector with itself is always the zero vector. Sometimes equation [latex]\nabla\times\nabla{(f)}=0[/latex] is simplified as [latex]\nabla\times\nabla=0[/latex].

theorem: curl test for a conservative field

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

Proof

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

[latex]_\blacksquare[/latex]

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

Example: testing whether a vector field is conservative

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

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

Similarly, if [latex]f[/latex] is a function of three variables then

[latex]\large{\text{div }(\nabla{f})=\nabla\cdot(\nabla{f})=f_{xx}+f_{yy}+f_{zz}}[/latex].

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

[latex]\large{\nabla^2f=0}[/latex].

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 [latex]f(x, y)=x^{2}+x-y[/latex] to be the potential function of an electrostatic field that is located in a region of [latex]\mathbb{R}^2[/latex] free of static charge?

try it

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