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The Divergence of a Vector Field
This section relies heavily on understanding vector fields from . We have separated the details of how the flux density of a region leads to the definition and understanding of the divergence ().
The Divergence of a Vector Field
This section relies heavily on understanding vector fields from . We have separated the details of how the flux density of a region leads to the definition and understanding of the divergence (). The second subsection is optional, but is worthwhile reading for students who are interested in a good conceptual understanding of how divergence is developed.
Introduction
As we saw in , there are many physical and theoretical representations for vector fields. A natural question is Where exactly is the vector field created? With the vector field in , imagine sketching a curve that follows the direction of the vector field by treating the vectors in the vector field as tangent vectors to your curve. No matter where you start, you should observe that the vector field decreases in strength as you move along the flow. We wish to understand (as a function of position), how much of the vector field is created (or destroyed) at a given location.
Exploration
In this preview activity, we will look at several two-dimensional vector fields and try to assess when the vector field has increased or decreased in strength over a given region. We begin with graphs of the three vector fields, \(\vF\), \(\vG\), and \(\vH\). Parts , , and ask you to answer the same three questions about the vector field and square illustrated in each of the figures. Part asks you to think further about the third vector field.
For each of the vector fields \(\vF\), \(\vG\), and \(\vH\) and the square centered on \(P_1\), \(P_2\), and \(P_3\) (respectively), which statement do you think best applies?
More of the vector field is going into the square than going out.
Less of the vector field is going into the square than going out.
The same amount of the vector field is going into the square as is going out.
For each of the vector fields \(\vF\), \(\vG\), and \(\vH\) (and corresponding square), does your answer to part suggest that the vector field is being created, destroyed, or is unchanging in strength inside the square? Write a sentence to explain your thinking for each vector field.
Would the answer to parts or change if you used a smaller square centered on \(P_1\), \(P_2\), and \(P_3\) for the corresponding vector fields? Write a sentence to explain your thinking for each vector field.
Thinking now only about the vector field \(\vH\), would your answers to parts , , or change if you considered squares around points \(P_4\), \(P_5\), or \(P_6\)? Write a couple of sentences to explain your thinking.
Definition of the Divergence of a Vector Field
We begin this subsection by stating a definition that captures analytically the ideas you reasoned about geometrically in . After the statement of the definition, we discuss what it means.
In other sources you may see the divergence written using a dot product as \(\divg(\vF) = \nabla\cdot \vF\). This notation is very compact and works well with the understanding that the del operator \(\nabla = \langle \frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}\rangle\) is a function that operates on other functions. However, this notation can also be confusing because of its emphasis on computation rather than conceptual understanding. In this text, we will not generally write the divergence using the del operator.
The divergence of a vector field is a scalar measurement at a point that measures how the strength of the vector field is changing as we look in small neighborhoods around our point. While will show the details for how the divergence is defined, we can make the following qualitative argument: If we are looking at how the strength of the vector field is changing in a small neighborhood of the point \((a,b)\), then we only need to look how fast the horizontal component is changing horizontally and how the vertical component is changing vertically. We make this measurement of changing strength of the vector field by measuring how much of a vector field flows into versus out of a small neighborhood of our point (as was done in ). When looking at a small neighborhood of the point (as shown by the square in ), only the change of the horizontal component of the vector field contributes to flow in or out on the sides and only the change in the vertical component contributes to flow in or out on the top and bottom.
Example
Let \(\vF=\langle x, y \rangle\). Then \[\begin{aligned}\end{aligned}\]. From our conceptual description above, \(\divg(\vF)\) being positive means that our vector field is increasing in strength. Since the divergence of \(\vF\) does not have a dependence on the input point, that means the vector field is increasing in strength, regardless of which point we consider. In the interactive element below, you can change the point you would like plotted and the size of the region around the point. You should see that regardless of what point you select or how small you make the region around the point, there will be more of the vector field flowing out of the region than in.
Condensed — the full section is in Boelkins et al., Active Calculus Multivariable.
Measuring the Change in Strength of a Vector Field
In this subsection, we examine the details of how to measure the density of the creation or destruction of the vector field using a classic calculus approach. Specifically, we will measure how the strength of the vector field changes in a region around a point. Next, using a limit, we examine what happens to our measurement as we shrink the region. Because vector fields change in a continuous fashion, the vector fields don't actually change at a single point. Rather, we will measure the density for the change in strength of the vector field using a limit.
We will develop all of our measurements in a two dimensional setting for now. However, our arguments can be applied to three (or more) dimensions. We start in the same fashion as in . Namely, we will look at how much of the vector field is going into or out of a square centered at a point \((a,b)\). For this development, we will consider a two-dimensional vector field given by \(\vF(x,y)=\langle{F_1(x,y),F_2(x,y)}\rangle\).
We can parametrize the top edge of the box by \(\vr_{\text{top}}(t) = \langle a+t,b+h\rangle\) with \(-h\leq t\leq h\). Similarly, the bottom, right, and left can be parametrized by \[\begin{aligned}\vr_{\text{bottom}}(t) \amp = \langle a+t,b-h\rangle \\ \vr_{\text{right}}(t) \amp = \langle a+h,b+t\rangle \\ \vr_{\text{left}}(t) \amp = \langle a-h,b+t\rangle\end{aligned}\] all of which use parameter values in \(-h\leq t\leq h\).
The amount of the vector field \(\vF\) that is created inside the square around the point \((a,b)\) can be measured by the net amount of the vector field coming into or going out of the square. The amount of vector flow that goes through each of the boundary segments can be measured by looking at just the orthogonal component of the vector field on each particular segment. For instance, on the top segment, the vertical component \(F_2\) determines how much of the vector field goes in or out of the square. Integrating just the vertical component \(F_2\) of the vector field \(\vF\) over the points on the top segment of our square will therefore measure how much of the vector field goes through the top of the square.
Condensed — the full section is in Boelkins et al., Active Calculus Multivariable.
Practice (1)
Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.
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- \(\vF=\langle{F_1,F_2,F_3}\rangle\)\[\begin{aligned}\end{aligned}\]\(\divg(\vG)=\vec{0}\)
Vector fields with a zero divergence everywhere in their domain are called divergence-free or incompressible vector fields. Which of the following vector fields are divergence-free?
- \(\vF=\langle{-y,z,x}\rangle\)
- \(\vF=\langle{\cos(yz),3xe^{z-x},6(x+y+z)^3}\rangle\)
- \(\vF=\langle{4xyz,y^2z,yz^2}\rangle\)
- \(\vF=\nabla f\)\(f\)\(x\)\(y\)\(z\)
- \(\vF_1=\langle{3(x-z)^2,2\cos(x)+3yz+y,-(z-1)^2+e^{xy}}\rangle\)\(\vF_1\)\(\divg(\vF_1)=0\)
- \(\vF_1\)
Paljasta vastaus
\[\begin{aligned}\divg(\vG) = \amp \frac{\partial}{\partial x}(\frac{\partial F_3}{\partial y}-\frac{\partial F_2}{\partial z})- \frac{\partial}{\partial y}(\frac{\partial F_3}{\partial x}-\frac{\partial F_1}{\partial z}) + \frac{\partial}{\partial z}(\frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y}) \\ = \amp (\frac{\partial^2 F_3}{\partial y \partial x}-\frac{\partial^2 F_2}{\partial z \partial x})- (\frac{\partial^2 F_3}{\partial x \partial y}-\frac{\partial^2 F_1}{\partial z \partial y}) + (\frac{\partial^2 F_2}{\partial x \partial z}-\frac{\partial^2 F_1}{\partial y \partial z}) \\ = \amp (\frac{\partial^2 F_3}{\partial y \partial x}-\frac{\partial^2 F_3}{\partial x \partial y}) + (\frac{\partial^2 F_2}{\partial x \partial z}-\frac{\partial^2 F_2}{\partial z \partial x}) + (\frac{\partial^2 F_1}{\partial z \partial y}-\frac{\partial^2 F_1}{\partial y \partial z})=0\end{aligned}\]
- \(\divg(\vF) = 0+0+0 =0\)
- \(\divg(\vF) = 0+0,18(x+y+z)^2\neq 0\)
- \(\divg(\vF) = 4yz+2yz+2yz \neq 0\)
- \(\divg(\vF) = \frac{\partial}{\partial x}(\frac{\partial f}{\partial x})+ \frac{\partial}{\partial y}(\frac{\partial f}{\partial y}) + \frac{\partial}{\partial z}(\frac{\partial f}{\partial z}) = \frac{\partial^2 f}{\partial x^2}+ \frac{\partial^2 f}{\partial y^2} + \frac{\partial^2 f}{\partial z^2}\)\(f(x,y,z)=x^2\)\(\divg(\vF) = \frac{\partial^2 f}{\partial x^2}+ \frac{\partial^2 f}{\partial y^2} + \frac{\partial^2 f}{\partial z^2} =2+0+0\)\(\vF=\nabla f\)
If \(\vF_1=\langle{3(x-z)^2,2\cos(x)+3yz+y,-(z-1)^2+e^{xy}}\rangle\), then \(\divg(\vF_1)= 6(x-z)+3z-2(z-1)=6x-5z+1\). Thus \(\divg(\vF_1) (-1,4,-1)=0\), but in general \(\vF_1\) is not divergence-free.
- \(\vF_1\)
Symbols used here
Derivative with respect to x, holding the other variables fixed.
Vector of partial derivatives; points uphill.
Ratios of sides in a right triangle; coordinates on the unit circle.
2.71828…, the base whose exponential is its own derivative.
Inequalities that allow equality; < and > exclude it.
Antiderivative (indefinite) or signed area from a to b (definite).
Integral over a region of the plane; integral around a closed curve.
A quantity with magnitude and direction; a column of numbers.
Σ u_i v_i; the length of v, √(v·v).
How to: The Divergence of a Vector Field
- How can you measure where a vector field's strength is increasing or decreasing?
- What does the divergence of a vector field measure and how can you visually estimate whether the divergence of a vector field is positive or negative?
Questions people ask
What is a partial derivative?
The ordinary derivative with respect to one variable while every other variable is frozen — the slope of the surface in one coordinate direction.
What does the gradient point at?
Uphill: the direction of steepest increase, with length equal to that steepest slope. It is perpendicular to the level curves.
Kokeile omaasi
Parts of this page are adapted from Boelkins et al., Active Calculus Multivariable (CC BY-SA 4.0). Condensed and re-explained here; errors are ours.
Lisää Multivariable Calculus
Functions of several variablesPartial derivatives and the gradientOptimisation in several variablesDouble and triple integralsVector fields, line integrals and the big theorems