# Divergence (in space)



My notes of the excellent lectures 28 and 29 by “Denis Auroux. 18.02 Multivariable Calculus. Fall 2007. Massachusetts Institute of Technology: MIT OpenCourseWare, https://ocw.mit.edu. License: Creative Commons BY-NC-SA.”

Divergence measures how much the flow is expanding. It singles out the stretching component of motion.

AKA as the Gauss-Green theorem. The 3D analogue to Green’s theorem for flux.

## Definition Closed surface $$S$$ and region $$D$$

If $$S$$ is a closed surface, enclosing a region $$D$$, oriented with $$\hat n$$ outwards and $$\vec F$$ defined and differentiable everywhere in $$D$$, then $$\newcommand{oiint}{\subset\!\!\supset\kern-1.65em\iint} \shaded{ \oiint_S\vec F\cdot d\vec S = \iiint_D\rm{div}(\vec F)\,dV } \label{eq:divthm}$$ The circle in the double integral reminds us that it must be a closed surface. The normal vectors always point outwards.

Let $$\vec F$$ be a vector field, where $$P,Q,R$$ are functions of $$x,y,z$$. \newcommand{pdv}{\frac{\partial #1}{\partial #2}} \shaded{ \begin{align*} \rm{div}\left(\vec F\right) &= \rm{div}\left(P\hat\imath+Q\hat\jmath+R\hat k\right) \\ &= \pdv{P}{x}+\pdv{Q}{y}+\pdv{R}{z} \\ &= P_x + Q_y + R_z \end{align*} } \nonumber

The divergence measures how much the flow is expanding. If you take a region of space, the total amount of water that flows out of it is the total amount of sources that you have in there minus the sinks.

### $$\nabla$$-notation

$$\nabla$$ “del” is a symbolic notation for the operator $$\newcommand{pdv}{\frac{\partial #1}{\partial #2}} \nabla = \left\langle\pdv{}{x}, \pdv{}{y}, \pdv{}{z}\right\rangle \nonumber$$

That makes \newcommand{pdv}{\frac{\partial #1}{\partial #2}} \begin{align*} \nabla f &= \left\langle\pdv{f}{x}, \pdv{f}{y}, \pdv{f}{z}\right\rangle & \text{gradient} \\[0.7em] \nabla\cdot\vec F &= \left\langle\pdv{}{x}, \pdv{}{y}, \pdv{}{z}\right\rangle \cdot \left\langle P,Q,R\right\rangle \\ &= \pdv{P}{x} + \pdv{Q}{y} + \pdv{R}{z} & \text{divergence} \end{align*}

This will come in very useful when we talk about curl in space.

## Physical interpretation

Let $$\vec F$$ be a vector field, where $$P,Q,R$$ are functions of $$x,y,z$$. \newcommand{oiint}{\subset\!\!\supset\kern-1.65em\iint} \newcommand{pdv}{\frac{\partial #1}{\partial #2}} \begin{align*} \oiint_S\vec F\cdot d\vec S &= \iiint_D\rm{div}(\vec F)\,dV \\[0.5em] \Rightarrow \oiint_S\left\langle P,Q,R\right\rangle &= \iiint_D\left(P_x+Q_y+R_z\right)\,dV \end{align*}

Claim that $$\rm{div}\left(\vec F\right) = \text{“source rate”} \nonumber$$

The amount of flux generated per unit volume. Think about an incompressible fluid, such as water. Incompressible fluid flow with velocity $$\vec F$$: (given mass occupies a fixed volume),

## Proof of divergence theorem

Simplifications

1. The dot-product can be seen a a sum of the different dimensions. We will proof one component: $$\newcommand{oiint}{\subset\!\!\supset\kern-1.65em\iint} \shaded{ \oiint_S\left\langle 0,0,R\right\rangle \cdot \hat n\,dS = \iiint_D R_z\,dV } \label{eq:simple1}$$ To proof the general case, we will prove the same thing for a vector field that has only an $$x$$ or $$y$$-component, and then sum the identities. \newcommand{oiint}{\subset\!\!\supset\kern-1.65em\iint} \begin{align*} \oiint_S\left\langle P,Q,R\right\rangle\hat n\,dS &= \oiint_S\left\langle P,0,0\right\rangle \cdot \hat n\,dS \\ &+ \oiint_S\left\langle 0,Q,0\right\rangle \cdot \hat n\,dS \\ &+ \oiint_S\left\langle 0,0,R\right\rangle \cdot \hat n\,dS \end{align*}
2. We will start with a vertically simple region. If region $$D$$ is vertically simple It lives above the shadow on the $$xy$$-plane, and is between two graphs $$z_1(x,y)$$ and $$z_2(x,y)$$.

$$\newcommand{oiint}{\subset\!\!\supset\kern-1.65em\iint} \oiint_S\left\langle 0,0,R\right\rangle \cdot \hat n\,dS = \iiint_D R_z\,dz \nonumber$$

We will compute the right- and left-hand sides of equation $$\eqref{eq:simple1}$$, and then compare them.

### Right-hand side

In right-hand side of equation $$\eqref{eq:simple1}$$, the range for $$z$$ is from the bottom face to the top face. \begin{align*} \iiint_D R_z\,dV &= \iint_U \underline{\int_{z_1(x,y)}^{z_2(x,y)} R_z\,dz}\,dx\,dy \end{align*}

When you integrate $$\newcommand{pdv}{\frac{\partial #1}{\partial #2}} R_z=\pdv{R}{z}$$ in respect to $$z$$, you just get $$R$$ back \begin{align} \iiint_D R_z\,dV &= \iint_U \Big[R(x,y,z)\Big]_{z_1(x,y)}^{z=z_2(x,y)}\,dx\,dy \nonumber \\ &= \iint_U \Big[R(x,y,z_2(x,y))-R(x,y,z_1(x,y))\Big]\,dx\,dy \label{eq:divthmproofleft} \\ \end{align}

### Left-hand side

The surface $$S$$ of the vertically simple region, consists of the top, bottom and the sides. Vertically simple region of $$S$$

We will calculate the flux on each side, and then sum them together \newcommand{oiint}{\subset\!\!\supset\kern-1.65em\iint} \begin{align} \oiint_S\left\langle 0,0,R\right\rangle \cdot \hat n\,dS &= \iint_\rm{top} + \iint_\rm{bottom} + \iint_\rm{sides} \label{eq:combine} \end{align}

#### Top

Recall equation $$\eqref{eq:fluxgraph}$$ for the flux of a graph

$$\hat n\,dS =\pm\left\langle -f_x,-f_y,1\right\rangle\,dx\,dy \nonumber$$

Apply this to the graph of $$z=z_2(x,y)$$ $$\newcommand{pdv}{\frac{\partial #1}{\partial #2}} \hat n\,dS = \left\langle -\pdv{z_2}{x}, -\pdv{z_2}{y}, 1 \right\rangle\,dx\,dy \nonumber$$

Do the dot-product with the vector field that only has a $$z$$-component and enter it in the double integral $$\eqref{eq:simple1}$$ \begin{align*} \iint_\text{top}\left\langle 0,0,R\right\rangle\cdot\hat n\,dS &= \iint_\text{top}\left\langle 0,0,R\right\rangle\cdot \left\langle -\pdv{z_2}{x}, -\pdv{z_2}{y}, 1 \right\rangle\\ &= \iint_\text{top} R\,dx\,dy \end{align*}

$$R$$ is $$R(x,y,z)$$. That means we can substitute the equation using the graph equation $$z=z_2(x,y)$$. As far as the region is concerned, it is above shadow $$U$$ \newcommand{pdv}{\frac{\partial #1}{\partial #2}} \begin{align} \iint_\rm{top} \left\langle 0,0,R\right\rangle\cdot\hat n\,dS &= \iint_U R(x,y,z_2(x,y))\,dx\,dy \label{eq:divthmproofright1} \\ \end{align}

#### Bottom

Similarly, apply equation $$\eqref{eq:fluxgraph}$$ to the graph of $$z=z_1(x,y)$$ $$\newcommand{pdv}{\frac{\partial #1}{\partial #2}} \hat n\,dS = \left\langle -\pdv{z_1}{x}, -\pdv{z_1}{y}, 1 \right\rangle \nonumber$$

Be care with the orientation of the normal vector. At the top surface, we want $$\hat n$$ pointing up, but at the bottom surface, we want $$\hat n$$ pointing down. Vertically simple region of $$S$$

Switch the orientation, so the normal vector points downwards \newcommand{oiint}{\subset\!\!\supset\kern-1.65em\iint} \newcommand{pdv}{\frac{\partial #1}{\partial #2}} \begin{align*} d\vec S &= \left\langle \pdv{z_1}{x},\pdv{z_1}{y}, -1\right\rangle\,dx\,dy \end{align*}

Do the dot-product with the vector field that only has a $$z$$-component and enter it in the double integral $$\eqref{eq:simple1}$$ \begin{align*} \iint_\text{bottom}\left\langle 0,0,R\right\rangle\cdot\hat n\,dS &= \iint_\text{bottom}\left\langle 0,0,R\right\rangle\cdot \left\langle\pdv{z_2}{x},\pdv{z_2}{y},-1 \right\rangle\\ &= \iint_\text{bottom} -R\,dx\,dy \end{align*}

$$R$$ is $$R(x,y,z)$$. That means we can substitute the equation using the graph equation $$z=z_2(x,y)$$. As far as the region is concerned, it is above shadow $$U$$ \newcommand{pdv}{\frac{\partial #1}{\partial #2}} \begin{align} \iint_\rm{bottom} \left\langle 0,0,R\right\rangle\cdot\hat n\,dS &= \iint_U -R(x,y,z_2(x,y))\,dx\,dy \label{eq:divthmproofright2} \\ \end{align}

#### Sides

$$\left\langle 0,0,R\right\rangle$$ is tangent to the sides $$\Longrightarrow$$ the flux through the sides $$=0$$. (that is why we took a vector field with only a $$z$$-component.)

In other words: the vector field $$\left\langle 0,0,R\right\rangle$$ is parallel to the $$z$$-axis, so at the sides there nothing going on.

#### Together

Combining equation $$\eqref{eq:divthmproofright1}$$ and $$\eqref{eq:divthmproofright2}$$ using $$\eqref{eq:combine}$$ \newcommand{oiint}{\subset\!\!\supset\kern-1.65em\iint} \begin{align} \oiint_S\left\langle 0,0,R\right\rangle \cdot \hat n\,dS = \iint_U &R(x,y,z_2(x,y))\,dx\,dy \nonumber \\ + \iint_U &-R(x,y,z_1(x,y))\,dx\,dy \label{eq:together} \end{align}

### Compare left- and right-hand side

Comparing $$\eqref{eq:divthmproofleft}$$ with $$\eqref{eq:together}$$ you indeed get the same formula \newcommand{pdv}{\frac{\partial #1}{\partial #2}} \newcommand{oiint}{\subset\!\!\supset\kern-1.65em\iint} \left\{ \begin{align*} \oiint_S\left\langle 0,0,R\right\rangle \cdot \hat n\,dS = \iint_U &\Big[R(x,y,z_2(x,y)) \\ &-R(x,y,z_1(x,y))\Big]\,dx\,dy \\ \iiint_D R_z\,dV = \iint_U &R(x,y,z_2(x,y))\,dx\,dy \\ + \iint_U &-R(x,y,z_1(x,y))\,dx\,dy \end{align*} \right.

Therefore, for a vertically simple region $$\newcommand{oiint}{\subset\!\!\supset\kern-1.65em\iint} \shaded{ \oiint_S\left\langle 0,0,R\right\rangle \cdot \hat n\,dS = \iiint_D R_z\,dV } \nonumber$$

### Generalization

1. If $$D$$ is not vertically simple, we cut it into vertically simple pieces. For instance, a solid doughnut is not vertically simple because above and below the hole makes two regions. However, if we slice it into four pieces, then each piece has a well defined top and bottom side (=vertically simple).
2. We could also do the same proof for the $$x$$ and $$y$$-component. Then sum them together to get the general case.
Redo the earlier example: Find the flux of $$\vec F=\left\langle x,y,z\right\rangle$$ through sphere of radius $$a$$ centered at the origin. The normal vector of the sphere points out radially from the origin. $$\vec F$$ and $$\hat n$$ point radially out
Apply the divergence theorem $$\eqref{eq:divthm}$$ \newcommand{oiint}{\subset\!\!\supset\kern-1.65em\iint} \begin{align*} \oiint_S\vec z\hat k\cdot d\vec S &= \iiint_D\rm{div}(z\hat k)\,dV \\ &= \iiint_D (0+0+1) \,dV \\ &= \underbrace{\iiint_D dV}_{\text{Volume}(D)} = \frac{4}{3}\pi a^3\\ \end{align*}