Divergence

Consider the vector field $\boldsymbol{F}$ defined in an area of space. The component of $\boldsymbol{F}$ is

$$\boldsymbol{F} = F_{1}\:\boldsymbol{i} + F_{2}\:\boldsymbol{j} + F_{3}\:\boldsymbol{k}$$

At this time, div $\boldsymbol{F}$ is called the divergence of the vector field and defined as follows.

Definition 3..2  

$${\rm div} \boldsymbol{F} = \frac{\partial F_{1}}{\partial x} + \frac{\partial F_{2}}{\partial y} + \frac{\partial F_{3}}{\partial z}$$

If we use the operator $\nabla$，we can write div $F=\nabla\cdot\boldsymbol{F}$ ．

Example 3..14  

Find the divergence of $${\boldsymbol{F} = 3xy\:\boldsymbol{i} + x^{2}y\:\boldsymbol{j} + y^{2}z\:\boldsymbol{k}}$$．

Answer

$${\rm div} \boldsymbol{F} = \frac{\partial F_{1}}{\partial x} + \frac{\partial F_{2}}{\partial y} + \frac{\partial F_{3}}{\partial z}$$

$$= 3y + x^{2} + y^{2}$$

．

Next, let's think about what the divergence of a vector field is, using actual physical phenomena. Here, we consider the movement of liquid, gas, etc. spreading in space. At this time, the velocity of the particles in that space is a vector field and ${\bf v}(x,y,z) = v_{1}\:\boldsymbol{i} + v_{2}\:\boldsymbol{j} + v_{3}\:\boldsymbol{k}$．Here, consider a Cartesian coordinate system with the point P in space as the origin, as shown in the figure . Imagine a small rectangular parallelepiped $\Delta x \Delta y \Delta z$ in a liquid．The area formed by $\Delta y \Delta z$ is denoted by $\Delta S_{yz}$.

Figure 3.6: divergenc

First, let's calculate the change in flow rate within a unit time of exiting from the surface of a rectangular parallelepiped. The $x$ component of the vector perpendicular to the plane of the rectangular parallelepiped of the fluid entering the rectangular parallelepiped at the point $(0,y,z)$ is $\rho v_{1}$, where $\rho$ is the density of the liquid. Therefore, the flow rate of the fluid flowing from the back surface within the unit time $\Delta t$ is

$$\rho v_{1}(0,y,z)\Delta S_{yz} \Delta t$$

Next, the $x$ component of the vector perpendicular to the plane of the rectangular parallelepiped of the fluid coming out of the rectangular parallelepiped is

$$\rho v_{1}(\Delta x,y,z) = \rho v_{1}(\Delta x, y, z) - \rho v_{1} (0,y,z) + \rho v_{1}(0,y,z)$$

$$= \frac{\rho v_{1}(\Delta x, y ,z) - \rho v_{1}(0,y,z)}{\Delta x} \Delta x + \rho v_{1}(0,y,z)$$

$$\approx \frac{\partial \rho v_{1}}{\partial x} \Delta x + \rho v_{1}(0,y,x)$$

Therefore, the flow rate of the fluid flowing out from the front surface is

$$\rho v_{1}(0,y,z)\Delta S_{yz} \Delta t + \left (\frac{\partial \rho v_{1}}{\partial x} \Delta x \right ) \Delta S_{yz} \Delta t$$

The same thing happens on the remaining four sides． Here, the surface integral of the vector field $\rho{\bf v}$ on six closed surfaces $\iint_{S}\rho{\ bf v} \cdot \boldsymbol{n}dS$ is ， As learned in the previous chapter, it represents the total flow velocity (total flow rate), so it can be approximated by adding all of these.

$$\iint_{S}\rho{\bf v}\cdot\boldsymbol{n}dS \approx \left (\frac{\p... ...al z} \right ) \Delta x \Delta y \Delta z = \nabla \cdot(\rho {\bf v})\Delta V$$

Here, the left side is the flow rate of the fluid flowing out from the surface $S$ of a small rectangular parallelepiped to the outside in 1 second. In other words, it is considered to be the flow rate of the fluid that springs out in one second. Therefore

The value at point P of $\nabla \cdot \rho {\bf v}$ is the volume density at point P of the volume of fluid that springs up in one second.．

This is the divergence of $\rho{\bf v}$ at P．Thus at P

$\nabla \cdot\rho {\bf v} > 0$ implies spill out.

$\nabla \cdot\rho {\bf v} < 0$ implies swallow.

$\nabla \cdot\rho {\bf v} = 0$ implies equilibrium.

Basic formula

For every vector field $\boldsymbol{F}$, $G$, and the scalar field $\phi$，the following formula holds.．

$$\nabla \cdot(\boldsymbol{F} +$$   $$\mbox{\boldmath$G$}$$$$) = \nabla \cdot\boldsymbol{F} + \nabla$$   $$\mbox{\boldmath$G$}$$

$$\nabla \cdot(\phi \boldsymbol{F}) = (\nabla \phi)\cdot\boldsymbol{F} + \phi \nabla \cdot\boldsymbol{F}$$

Proof

$$\nabla \cdot(\boldsymbol{F} + \mbox{\boldmath$G$} ) = (\boldsymbol{i} \frac{\partial}{\partial x} + \boldsymbol{j} \fra... ...\partial y} + \boldsymbol{k}\frac{\partial}{\partial z}) \cdot(\boldsymbol{F} + \mbox{\boldmath$G$} )$$

$$= (\boldsymbol{i} \frac{\partial}{\partial x} + \boldsymbol{j} \fra... ... \frac{\partial}{\partial y} + \boldsymbol{k}\frac{\partial}{\partial z}) \cdot \mbox{\boldmath$G$}$$

$$= \nabla \cdot\boldsymbol{F} + \nabla \mbox{\boldmath$G$}$$

$$\nabla \cdot(\phi \boldsymbol{F}) = (\boldsymbol{i} \frac{\partial}{\partial x} + \boldsymbol{j} \fra... ...rtial y} + \boldsymbol{k}\cdot\frac{\partial (\phi \boldsymbol{F})}{\partial z}$$

$$= \boldsymbol{i}\cdot(\frac{\partial \phi}{\partial x} \boldsymbol{... ...i}{\partial z} \boldsymbol{F} + \phi \frac{\partial\boldsymbol{F}}{\partial z})$$

$$= (\nabla \phi)\cdot\boldsymbol{F} + \phi \nabla \cdot\boldsymbol{F}$$

Example 3..15  

For $\phi = 3x^2 - yz$, $\boldsymbol{F} = 3xyz^2\:\boldsymbol{i} + 2xy^3\:\boldsymbol{j} -x^2yz\:\boldsymbol{k}$，find the following scalar.
(1) $\nabla \cdot\boldsymbol{F}$(2) $\boldsymbol{F} \cdot\nabla \phi$(3) $\nabla \cdot(\phi \boldsymbol{F})$ Answer (1)

$$\nabla \cdot\boldsymbol{F} = \frac{\partial}{\partial x}(3xyz^2) ... ...\partial y}(2xy^3) + \frac{\partial}{\partial z}(-x^2yz) = 3yz^2 + 6xy^2 - x^2y$$

(2) $\nabla \phi = 6x\:\boldsymbol{i} - z\:\boldsymbol{j} - y\:\boldsymbol{k}$ implies

$$\boldsymbol{F} \cdot\nabla \phi = (3xyz^2)(6x) + (2xy^3)(-z) + (-x^2yz)(-y) = 18x^2yz^2 - 2xy^3z + x^2y^2z$$

Question 3..7  

find (3)．

Example 3..16  

Let $\boldsymbol{r} = x\boldsymbol{i} + y\boldsymbol{j} + z\boldsymbol{k},\ r = \vert\boldsymbol{r}\vert$．Prove the following:

$$(1)\ \nabla \cdot\boldsymbol{r} = 3,\hskip 1cm (2)\ \nabla \cdot(... ...ol{r}) = (n+3)r^{n}\hskip 1cm (3)\ \nabla \cdot(\frac{\boldsymbol{r}}{r^3}) = 0$$

Answer (1)

$$\nabla \cdot\boldsymbol{r} = (\frac{\partial}{\partial x}\boldsym... ...l{k})\cdot(x\boldsymbol{i} + y\boldsymbol{j} + z\boldsymbol{k}) = 1 + 1 + 1 = 3$$

(2) By the differentiation of composite functions,，

$$\nabla \cdot(r^{n}\boldsymbol{r}) = (\nabla r^{n}) \cdot\boldsymb... ...boldsymbol{r}}{r} \cdot\boldsymbol{r} + r^{n}(3) = nr^{n} + 3r^{n} = (n+3)r^{n}$$

(3)

$$\nabla \cdot(\frac{\boldsymbol{r}}{r^3}) = \nabla \cdot r^{-3}\bo... ...ymbol{r} = -3r^{-4}\frac{\boldsymbol{r}}{r} \cdot\boldsymbol{r} + r^{-3}(3) = 0$$

Laplacian

If $\boldsymbol{F}$ is conservative, then $\rm {div}\boldsymbol{F}$ can be expressed by

$$\rm {div}\boldsymbol{F} = \nabla \cdot\boldsymbol{F} = \nabla \cdot\nabla \phi$$

$$= \left(\frac{\partial}{\partial x}\:\boldsymbol{i} + \frac{\partia... ...l y}\:\boldsymbol{j} + \frac{\partial \phi}{\partial z}\:\boldsymbol{k} \right)$$

$$= \frac{\partial^2 \phi }{\partial x^2 }\:\boldsymbol{i} + \frac{\p... ...ial y^2}\:\boldsymbol{j} + \frac{\partial^2 \phi}{\partial z^2}\:\boldsymbol{k}$$

This is called Laplacian of the scalar field $f$ and denoted by $\nabla^2 \phi$ or $\Delta \phi$．Here，the inner product of the operators $\nabla \cdot\nabla$ is represented by $\nabla^2$. Then

$$\nabla^2 = \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2}$$

Also, the partial differentiation equation

$$\nabla^2 \phi = \frac{\partial^2 \phi }{\partial x^2 } + \frac{\partial^2 \phi}{\partial y^2} + \frac{\partial^2 \phi}{\partial z^2} = 0$$

is called Laplace equation and the solution to that equation $\phi$ is calledharmonic function ．

Example 3..17  

Find $\phi = 3x^2 y - y^3z^2$のとき， $\nabla^2 \phi$． Answer

$$\nabla^2 \phi = \frac{\partial^2}{\partial x^2}(3x^2 y - y^3z^2) ... ... y^3z^2) + \frac{\partial^2}{\partial z^2}(3x^2 y - y^3z^2) = 6y - 6yz^2 - 2y^3$$

Question 3..8  

Let $\boldsymbol{r} = x\boldsymbol{i} + y\boldsymbol{j} + z \boldsymbol{k}$, $r= \vert\boldsymbol{r}\vert$． Find the followings.

(1) $\nabla^2 r^{n} = n(n+1)r^{n-2}$ $(2)\ \nabla^2 (\frac{1}{r}) = 0$

Example 3..18  

Let $\boldsymbol{r} = x\boldsymbol{i} + y\boldsymbol{j} + z \boldsymbol{k}$, $r= \vert\boldsymbol{r}\vert$．

(1) Prove that $${\nabla^2 f(r) = \frac{d^{2}f}{dr^2} + \frac{2}{r}\frac{df}{dr}}$$よ．

(2) Find $f(r)$ so that $${\nabla^2 f(r) = 0}$$.．

Answer (1) $w = f(r), r = r(x,y,z)$implies

$$\nabla f(r) = \frac{d f}{dr}(\frac{\partial r}{\partial x}\boldsy... ...bol{j} + \frac{\partial r}{\partial z}\boldsymbol{k}) = \frac{d f}{dr} \nabla r$$

Thus，

$$\nabla^2 f(r) = \nabla \cdot(\frac{d f}{dr} \nabla r) = \nabla (\frac{d f}{dr}) \cdot\nabla r + \frac{d f}{dr} \nabla \cdot\nabla r$$

Note that $\nabla r \cdot\nabla r = \frac{\boldsymbol{r}}{r} \cdot\frac{\boldsymbol{r}}{r} = 1$. Also， $\nabla \cdot\nabla r = \nabla \cdot\frac{\boldsymbol{r}}{r} = \nabla \cdot(r^{-... ...symbol{r} = -r^{-2}\boldsymbol{r} \cdot\boldsymbol{r} + r^{-1}(3) = \frac{2}{r}$. Therefore，

$$\nabla^2 f(r) = \frac{d^{2}f}{dr^2} + \frac{2}{r}\frac{df}{dr}$$

(2) Since $\nabla^2 f(r) = 0$，we have

$$\frac{d^{2}f}{dr^2} + \frac{2}{r}\frac{df}{dr} = 0$$

Note that if we let $w = \frac{d f}{dr}$，then $\frac{d w}{dr} = - \frac{2}{r}w$ and it is separable．Thus， $\frac{d w}{w} = -\frac{2}{r}dr$ and integrate both sides, we have

$\log\vert w\vert = -2\log\vert r\vert + c$. Thus， $w = cr^{-2}$. $w = \frac{d f}{dr}$ and $\frac{d f}{dr } = cr^{-2}$．Therefore， $f = c_{0}r^{-1} + c_{1}$.

Exercise3.5

fundamental formula Let $\boldsymbol{r} = x\boldsymbol{i} + y\boldsymbol{j} + z\boldsymbol{k},\ r = \sqrt{x^2 + y^2 + z^2}$. Then
(1) $\nabla r = \frac{\boldsymbol{r}}{r}$
(2) $\nabla r^{n} = nr^{n-1}\nabla r = nr^{n-2}\boldsymbol{r}$
(3) $\nabla \cdot(\phi\boldsymbol{A})= (\nabla \phi) \cdot\boldsymbol{A} + \phi \nabla \cdot\boldsymbol{A}$

1.

Find the followings.．

(1) $\nabla \cdot(2x^2 z\boldsymbol{i} - xy^2 z \boldsymbol{j} + 3yz^2 \boldsymbol{k})$(2) $\nabla^2(3x^2 z - y^2 z^3 + 4x^2 y)$

(3) $\nabla(\nabla \cdot\boldsymbol{F}), \ \boldsymbol{F} = (3x^2 y - z)\boldsymbol{i} + (xz^{3} + y^{4})\boldsymbol{j} - 2x^2 z^2 \boldsymbol{k}$

2.

Let $\boldsymbol{r} = x\boldsymbol{i} + y\boldsymbol{j} + z\boldsymbol{k},\ r = \vert\boldsymbol{r}\vert$．Find the following scalar．

(1) $\nabla \cdot(r\nabla r^{-3})$(2) $\nabla^{2}\left\{\nabla\cdot\left(\frac{\boldsymbol{r}}{r^2}\right)\right\}$

3.

Let $\boldsymbol{r} = x\boldsymbol{i} + y\boldsymbol{j} + z\boldsymbol{k},\ {\bf w}$ be a constant vector. Then show that ．

$$\nabla \cdot({\bf w} \times \boldsymbol{r}) = 0$$

4.

For the scalar fields $\phi, \psi$, prove the followings. (1) $\nabla^2(\phi \psi) = \phi \nabla^2 \psi + 2(\nabla \phi)\cdot(\nabla \psi) + \psi \nabla^2 \phi$

(2) $\nabla \cdot(\phi \nabla \psi) = (\nabla \phi) \cdot(\nabla \psi) + \phi \nabla^2 \psi$

(3) $\nabla \cdot(\phi \nabla \psi - \psi \nabla \phi) = \phi \nabla^2 \psi - \psi \nabla^2 \phi$

5.

Find $\phi = \phi(x,y,z)$ which satisfies $\phi = \phi(x,y,z)$ $\nabla \phi = 2xyz^3 \boldsymbol{i} + x^2 z^3 \boldsymbol{j} + 3x^2 y z^2 \boldsymbol{k}$．provided $\phi(1,-2,2) =4$．

6.

For scalar fields $U,V$, show the following.

$$\nabla \cdot\{(\nabla U) \times (\nabla V)\} = 0$$

7.

For the curve $\boldsymbol{r} = \boldsymbol{r}(t) = x(t)\boldsymbol{i} + y(t)\boldsymbol{j} + z(t)\boldsymbol{k}$ and the scalar field $\phi = \phi(x,y,z)$，show that the derivative $\frac{d\phi(x(t),y(t),z(t))}{dt}$ of $\phi$ along the curve is equal to $\frac{d\boldsymbol{r}}{dt}\cdot\nabla \phi$.

8.

Prove that the derivative of $\phi(t)$ which is composed by putting $x = x(t), y = y(t), z = z(t)$ into $\phi(x,y,z,t)$ is $\frac{d\phi}{dt} = \frac{\partial \phi}{\partial t} + \frac{d\boldsymbol{r}}{dt} \cdot\nabla \phi$．provided， $\boldsymbol{r} = x\boldsymbol{i} + y\boldsymbol{j} + z \boldsymbol{k}$.