Divergence and Green’s Theorem

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Divergence measures the rate field vectors are expanding at a point.

While the gradient and curl are the fundamental “derivatives” in two dimensions, there is another useful measurement we can make. It is called divergence. It measures the rate field vectors are “expanding” at a given point.

The divergence of a vector field

Let’s state the definition:

Given a vector field $\vec {F}:\R ^n\to \R ^n$ , where $\vec {F}(x_1,\dots ,x_n) = \begin {bmatrix} M_1(x_1,\dots ,x_n)\\ M_2(x_1,\dots ,x_n)\\ \vdots \\ M_n(x_1,\dots ,x_n) \end {bmatrix}$ the divergence is given by $\divergence \vec F =\pp [M_1]{x_1} + \pp [M_2]{x_2} + \dots + \pp [M_n]{x_n}.$ Some authors use the notation $\mathop {\mathrm {div}}\vec {F}$ for the divergence of a vector field.

Is the divergence of a vector field a scalar or a vector?

vector. scalar. neither a vector nor a scalar.

The divergence is a number that tells you how much the field is expanding at a point. However, no directional information is given.

Consider the vector field $\vec {F}(x,y) = \vector {x^2y^3,ye^x}$ . Compute: $\divergence \vec {F} \begin {prompt} = \answer {2xy^3 + e^x} \end {prompt}$

Consider the vector field $\vec {F}(x,y,z) = \vector {\sin (xy),\sin (x+y+z),\sin (yz}$ . Compute: $\divergence \vec {F} \begin {prompt} = \answer {y\cos (xy) + \cos (x+y+z) + y\cos (yz)} \end {prompt}$

What does divergence measure?

As we’ve already said, divergence measures the rate field vectors are expanding at a point. To be more explicit, the divergence measures how the magnitude of the field vectors change as you move in the direction of the field vectors:

And:

The most obvious example of a vector field with nonzero divergence is $\vec {F}(x,y)= \vector {x,y}$ :

On the other hand, recall that a radial vector field is a field of the form $\vec {F}:\R ^n\to \R ^n$ where $\vec {F}(\vec {x}) = \frac {\pm \vec {x}}{|\vec {x}|^p}$ where $p$ is a real number. The divergence of these vector fields can be surprising.

Compute the divergence of $\vec {F}(x,y) =\frac {\vector {x,y}}{|\vector {x,y}|^p}$ .

First note $\begin{align*} \pp{x} \frac{x}{(\sqrt{x^2+y^2})^p} &= \pp{x} \frac{x}{(x^2+y^2)^{p/2}} \\ &= \answer[given]{\frac{(x^2+y^2)^{p/2}-px^2 (x^2+y^2)^{p/2-1}}{(x^2+y^2)^p}} \end{align*}$ and next note $\begin{align*} \pp{y} \frac{y}{(\sqrt{x^2+y^2})^p} &= \pp{y} \frac{y}{(x^2+y^2)^{p/2}} \\ &= \answer[given]{\frac{(x^2+y^2)^{p/2}-py^2 (x^2+y^2)^{p/2-1}}{(x^2+y^2)^p}} \end{align*}$ so $\begin{align*} \divergence\vec{F}(x,y) = \frac{2-p}{|\vector{x,y}|^p} \end{align*}$

Compute the divergence of $\vec {F} =\frac {\vector {x,y,z}}{|\vector {x,y,z}|^p}$ . $\divergence \vec {F}(x,y,z) = \frac {\answer [given]{3-p}}{|\vector {x,y,z}|^p}$

Now we will see a radial vector field with zero divergence.

Consider the vector field $\vec {F}(x,y) = \frac {\vector {x,y}}{|\vector {x,y}|^2}$ :

Compute $\divergence \vec {F}$ .

Since this is a radial field, we know its divergence is $\frac {2-\answer [given]{2}}{|\vector {x,y}|^p} = \answer [given]{0}$ when $\vector {x,y}\ne \vec {0}$ .

Now let’s see a radial vector field with negative divergence.

Consider the vector field $\vec {F}(x,y) = \frac {\vector {x,y}}{|\vector {x,y}|^3}$ :

Compute $\divergence \vec {F}$ .

Since this is a radial field, we know its divergence is $\frac {2-\answer [given]{3}}{|\vector {x,y}|^p} < 0$ when $\vector {x,y}\ne \vec {0}$ .

Measuring flow across a curve

Let $\vec {F}:\R ^2\to \R ^2$ be a vector field, $\vec {p}:\R \to \R ^2$ be a smooth vector valued function tracing a curve $C$ exactly once as $t$ runs from $a$ to $b$ ,

$\begin{align*} \vec{F}(x,y) &= \vector{M(x,y), N(x,y)}\\ \vec{p}(t) &= \vector{x(t),y(t)}. \end{align*}$ Recall that the line integral $\int _C \vec {F}\dotp \d \vec {p}$ measures the accumulated flow of a vector field along a curve. We see this because $\vec {F}\dotp \vec {p}'$ measures how “aligned” field vectors are with the direction of the path $\vec {p}$ . On the other hand, if we set $\vec {n}(t) = \vector {y(t),-x(t)}$ then for any given value of $t$ , $\vec {n}'(t)$ is a vector that is orthogonal to $\vec {p}'(t)$ . Moreover, given a closed curve, where $\vec {p}(t)$ is parameterized with the interior on the left, $\vec {n}'(t)$ points outward. Below we see a curve $\vec {p}(t)$ along with some tangent vectors $\vec {p}'(t)$ and some outward normal vectors $\vec {n}'(t)$ :

Since $\vec {F}\dotp \vec {n}'$ measures how “aligned” field vectors are with vectors orthogonal to the direction of the path, the integral $\begin{align*} \oint_C \vec{F}\dotp \d \vec{n} &= \oint_C \vector{M,N}\dotp\vector{\d y,-\d x}\\ &= \oint_C - N(x,y)\d x + M(x,y)\d y \end{align*}$ measures the flow of a vector field across a curve. Some folks call this a flux integral. Since $\d x = x'(t)\d t$ and $\d y = y'(t)\d t$ , we may write $\oint _C \vec {F}\dotp \d \vec {n}$ as $\begin{align*} &= \int_a^b \left(-N(x(t),y(t))\cdot x'(t) + M(x(t),y(t))\cdot y'(t)\right) \d t\\ &= \int_a^b \vector{M(x(t),y(t)),N(x(t),y(t))}\dotp\vector{y'(t), -x'(t)} \d t \end{align*}$

Consider the following vector field $\vec {F}$ and curve $C$ parameterized by $\vec {p}(t)$ :

Do you expect $\oint _C\vec {F} \dotp \d \vec {n}$ to be positive, zero, or negative?

positive zero negative

Consider the following vector field $\vec {F}$ and curve $C$ parameterized by $\vec {p}(t)$ :

Do you expect $\oint _C\vec {F} \dotp \d \vec {n}$ to be positive, zero, or negative?

positive zero negative

Consider the following vector field $\vec {F}$ and curve $C$ parameterized by $\vec {p}(t)$ :

Do you expect $\oint _C\vec {F} \dotp \d \vec {n}$ to be positive, zero, or negative?

positive zero negative

With our next example, we’ll get our hands dirty.

Below we see a closed curve $C$ along with some representative field vectors of a vector field $\vec {F}$ :

Setting $\d x =1$ or $-1$ (depending on the direction of the curve), estimate: $\oint _C \vec {F}\dotp \d \vec {n}$

If $\vec {F}(x,y)= \vector {M(x,y),N(x,y)}$ , we have that: $\oint _C \vec {F}\dotp \d \vec {n} = \oint _C \left (-N(x,y)\d x + M(x,y)\d y\right )$ We know that $|\d x|= \answer [given]{1}$ , and $\d y = y'(x) \d x$ . Now we’ll compute $\sum \left (-N(x,y) \d x + M(x,y) \d y\right )$ for each edge of the triangle above. For the bottom edge $\d x = \answer [given]{1}$ and $\d y = \answer [given]{0}$ . So we have: $\sum _{\mathrm {bottom}} \left (-N(x,y) \d x + M(x,y) \d y\right ) = \answer [given]{6}$ Along the right edge, $\d x = \answer [given]{-1}$ and $\d y = \answer [given]{1/2}$ . So $\sum _{\mathrm {right}} \left (-N(x,y) \d x + M(x,y) \d y\right ) = \answer [given]{6}$ Finally, along the left edge, $\d x = \answer [given]{-1}$ and $\d y = \answer [given]{-1}$ . So $\sum _{\mathrm {left}} \left (-N(x,y) \d x + M(x,y) \d y\right ) = \answer [given]{2}.$ Hence we estimate $\oint _C \vec {F}\dotp \d \vec {n} \approx \answer [given]{14}.$

Let $\vec {F}(x,y) = \vector {x,y}$ and let $C$ be the unit circle centered at the origin. Compute $\oint _C \vec {F}\dotp \d \vec {n}$

The path $C$ can be parameterized by $\begin{align*} x(\theta) &= \cos(\theta)\\ y(\theta) &= \sin(\theta) \end{align*}$ with $0\le \theta \le 2\pi$ . To compute the integral, write with me $\begin{align*} \oint_C \vec{F}\dotp\d\vec{n} &= \int_0^{2\pi} F(x(\theta),y(\theta))\dotp \vector{y'(\theta),-x'(\theta)}\d \theta\\ &= \int_0^{2\pi} \vector{\answer[given]{\cos(\theta)},\answer[given]{\sin(\theta)}}\dotp \vector{\answer[given]{\cos(\theta)},\answer[given]{\sin(\theta)}}\d \theta\\ &= \int_0^{2\pi}\left(\cos^2(\theta)+\sin^2(\theta)\right)\d \theta\\ &= \int_0^{2\pi} 1\d \theta\\ &=\answer[given]{2\pi}. \end{align*}$

Connections to Green’s Theorem

Finally, note that if $\vec {F}=\vector {M,N}$ , then:

$\begin{align*} \divergence \vec{F} &=\pp[M]{x} + \pp[N]{y}\\ &=\curl\vector{-N,M} \end{align*}$ We also see that $\begin{align*} \int_C \vector{-N,M}\dotp \d\vec{p} &= \int_C \left(-N\d x+ M \d y\right)\\ &=\int_C \vec{F}\dotp\d\vec{n} \end{align*}$ this leads us to the flux form of Green’s Theorem:

Green’s Theorem If the components of $\vec {F}:\R ^2\to \R ^2$ have continuous partial derivatives and $C$ is a boundary of a closed region $R$ and $\vec {p}(t) = \vector {x(t),y(t)}$ parameterizes $C$ in a counterclockwise direction with the interior on the left, and $\vec {n}(t) = \vector {y(t),-x(t)}$ , then $\iint _R \divergence \vec {F}\d A = \oint _C \vec {F}\dotp \d \vec {n}$

Let $\vec {F}$ be a vector field with $\divergence \vec {F} = 0$ . Compute: $\oint _C \vec {F}\dotp \d \vec {n} \begin {prompt} =\answer {0} \end {prompt}$

Suppose that the divergence of a vector field $\vec {F}:\R ^2\to \R ^2$ is constant, $\divergence \vec {F} = -4$ .

If $\int _{C_1} \vec {F} \dotp \d \vec {n} = 20$ estimate: $\int _{C_2} \vec {F}\dotp \d \vec {n} \begin {prompt} = \answer {-108} \end {prompt}$

Use Green’s Theorem.