Divergence

If \({\bf F} = F_1 {\bf i} + F_2 {\bf j}+ F_3 {\bf k}\), the divergence of \({\bf F}\) is the scalar field \[ \hbox{div } {\bf F} = \nabla\, {\cdot}\, {\bf F}= \frac{\partial F_1}{\partial x} + \frac{\partial F_2}{\partial y}+ \frac{\partial F_3}{\partial z}. \]

example 1

Compute the divergence of \[ {\bf F} = x^2 y {\bf i} + z {\bf j} + xy z {\bf k}. \]

solution \[ \hbox{div } {\bf F} = \frac{\partial}{\partial x} ( x^ 2 y) + \frac{\partial}{\partial y} (z) + \frac{\partial}{\partial z} (xyz)=2 xy + 0 + xy = 3xy \]

For further discussion and the direct proof of this interpretation of the divergence, click here.

\(\bf Flows \ of \ Vector \ Fields.\) It is convenient to give the unique solution through a given point at time \(0\) a special notation: \[ \phi (\mathbf{x}, t) = \left\{ \begin{array}{l}\mbox{the position of the point on the flow line}\\ \mbox{through the point \(\mathbf{x}\) after time \(t\) has elapsed}. \end{array} \right\}. \] With \(\mathbf{x}\) as the initial condition, follow along the flow line for a time period \(t\) until the new position \(\phi(\mathbf{x}, t)\) is reached (see the Figure below).

The definition of the flow \( \phi(\mathbf{x}, t) \) of \(\mathbf{F}\).

Alternatively, \(\phi (\mathbf{x}, t) \) is defined by: \begin{equation} \label {flowequation} \left.\begin{array}{c} \displaystyle{\frac{\partial }{\partial t}} \phi(\mathbf{x}, t) = \mathbf{F} (\phi(\mathbf{x}, t)) \\ \phi(\mathbf{x}, 0) = \mathbf{x} \end{array} \right\}.\tag{1} \end{equation} We call the mapping \(\phi\), which is regarded as the function of the variables \(\mathbf{x}\) and \(t\), the \({\bf flow}\) of \(\mathbf{F}\). Let \(\mathbf{D}_{\mathbf{x}}\) denote differentiation with respect to \(\mathbf{x}\), holding \(t\) fixed. It is proved in courses on differential equations that \(\phi\) is, in fact, a differentiable function of \(\mathbf{x}\). By differentiating equation (1) with respect to \(\mathbf{x}\), we get \[ \mathbf{D}_{\mathbf{x}} \frac{\partial }{\partial t} \phi(\mathbf{x}, t) = \mathbf{D}_{\mathbf{x}}[\mathbf{F}( \phi (\mathbf{x}, t))] . \] The equality of mixed partial derivatives may be used on the left-hand side of this equation and the chain rule applied to the right-hand side, yielding \begin{equation}\label{equation2} \frac{\partial }{\partial t}\mathbf{D}_{\mathbf{x}} \phi (\mathbf{x}, t) = \mathbf{D}\mathbf{F}(\phi (\mathbf{x}, t))\mathbf{D}_{\mathbf{x}} \phi (\mathbf{x}, t) ,\tag{2} \end{equation} where \(\mathbf{D}\mathbf{F}(\phi (\mathbf{x}, t))\) denotes the derivative of \(\mathbf{F}\) evaluated \( \phi(\mathbf{x}, t)\). Equation (2), a linear differential equation for \( \mathbf{D}_{\mathbf{x}} \phi (\mathbf{x}, t)\) is called the {\bfi equation of the first variation}. It will be useful in our discussion of divergence in the next section. Both \(\mathbf{D}_{\mathbf{x}} \mathbf{F}(\phi)\) and \(\mathbf{D}_{\mathbf{x}} \phi \) are \(3 \times 3\) matrices since \(\mathbf{F}\) and \(\phi\) take values in \(\mathbb{R}^3\) and are differentiated with respect to \( \mathbf{x} \in \mathbb{R}^3\); for vector fields in the plane, they would be \(2 \times 2\) matrices.

\(\bf The \ Geometry \ of \ the \ Divergence.\) We now study the geometric meaning of the divergence in more detail. This discussion depends on the concept of the flow \(\phi (\mathbf{x}, t)\) of a vector field \(\mathbf{F}\) given in the preceding paragraph.

Fix a point \(\mathbf{x}\) and consider the three standard basis vectors \(\mathbf{i}, \mathbf{j}, \mathbf{k}\) emanating from \(\mathbf{x}\). Let \(\varepsilon \gt 0\) be small and consider the basis vectors \( \mathbf{v}_1 = \varepsilon \mathbf{i} , \mathbf{v}_2 = \varepsilon \mathbf{j}, \mathbf{v}_3 = \varepsilon \mathbf{k}\), also emanating from \(\mathbf{x}\). These vectors span a parallelepiped \(P(0)\). As time increases or decreases, the flow \(\phi(\mathbf{x}, t)\) carries \(P(0)\) into some object. For \({\it fixed}\) time, \(\phi\) is a differentiable function of \(\mathbf{x}\) (that is, \(\phi\) is a differentiable map of \(\mathbb{R}^3\) to \(\mathbb{R}^3\)). When \(\varepsilon \) is small, the image of \(P(0)\) under \(\phi\) can be approximated by its image under the derivative of \( \phi\) with respect to \(\mathbf{x}\). Recall that if \(\mathbf{v}\) is a vector based at a point \(P_1\) and ending at \(P_2\), so \(\mathbf{v} = P_2 - P_1\), then \(\phi(P_2, t) - \phi (P_1, t) \approx \mathbf{D}_{\mathbf{x}} \phi (\mathbf{x}, t) \cdot \mathbf{v}\). Thus for fixed time and small positive \( \varepsilon, P(0)\) is approximately carried into a parallelepiped spanned by the vectors \( \mathbf{v}_1 (t), \mathbf{v}_2 (t), \mathbf{v}_3 (t)\) given by \begin{equation}\label{equation3} \left. \begin{array}{c} \mathbf{v}_1(t) = \mathbf{D}_{\mathbf{x}} \phi (\mathbf{x}, t) \cdot \mathbf{v}_1\\ \mathbf{v}_2 (t) = \mathbf{D}_{\mathbf{x}} \phi (\mathbf{x}, t) \cdot \mathbf{v}_2\\ \mathbf{v}_3 (t) = \mathbf{D}_{\mathbf{x}} \phi (\mathbf{x}, t) \cdot \mathbf{v}_3 \end{array} \right\}.\tag{3} \end{equation} Since \(\phi( \mathbf{x}, 0) = \mathbf{x}\) for all \(\mathbf{x}\), it follows that \(\mathbf{v}_1(0) = \mathbf{v}_1, \mathbf{v}_2(0) = \mathbf{v}_2\), and \(\mathbf{v}_3(0) = \mathbf{v}_3\). In summary, the vectors \( \mathbf{v}_1 (t), \mathbf{v}_2(t), \mathbf{v}_3 (t)\) span a parallelepiped \(P(t) \) that moves in time (see the Figure below).

The moving basis \(\mathbf{v}_1(t), \mathbf{v}_2(t), \mathbf{v}_3(t)\) and the associated parallepiped.

Let the volume of \(P(t)\) be denoted by \(\mathcal{V}(t)\). The main geometric meaning of divergence is given by the following theorem.

\({\bf Theorem.}\) \[\operatorname{div} \mathbf{F}(\mathbf{x}) \left. = \frac{1}{\mathcal{V}(0)} \frac{d}{dt} \mathcal{V} (t) \right|_{t = 0} .\]

\(\bf proof\) By equation (2) of the previous paragraph, \begin{equation}\label{equation4} \frac{d}{dt} \mathbf{v}_i (t) = \mathbf{D} \mathbf{F} (\phi(\mathbf{x}, t)) \cdot (\mathbf{D}_{\mathbf{x}} \phi (\mathbf{x}, t) \cdot \mathbf{v}_i)\tag{4} \end{equation} for \(i = 1, 2, 3\). Since \( \phi(\mathbf{x}, 0) = \mathbf{x}\), it follows that \(\mathbf{D}_{\mathbf{x}} \phi (\mathbf{x}, 0)\) is the identity matrix, so that evaluation at \(t=0\) gives \[ \left.\frac{d}{dt} \mathbf{v}_i (t) \right|_{t = 0} = \mathbf{D} \mathbf{F}(\mathbf{x}) \cdot \mathbf{v}_i. \] The volume \(\mathcal{V} (t)\) is given by the triple product: \[ \mathcal{V}(t) = \mathbf{v}_1 (t) \cdot [\mathbf{v}_2(t) \times \mathbf{v}_3(t)]. \] Using the differentiation rules of \S4.1 and the identities \[ \mathbf{v}_1 \cdot [\mathbf{v}_2 \times \mathbf{v}_3 ] = \mathbf{v}_2 \cdot[\mathbf{v}_3 \times \mathbf{v}_1] = \mathbf{v}_3 \cdot [ \mathbf{v}_1 \times \mathbf{v}_2], \] equation (\ref{equation4}) gives \begin{eqnarray*} \frac{d \mathcal{V}}{dt} &= &\frac{d \mathbf{v}_1}{dt} \cdot [ \mathbf{v}_2 (t) \times \mathbf{v}_3(t)] + \mathbf{v}_1 (t) \cdot \left[ \frac{d \mathbf{v}_2}{dt} \times \mathbf{v}_3(t)\right] + \mathbf{v}_1(t) \cdot \left[ \mathbf{v}_2 (t) \times \frac{d \mathbf{v}_3}{dt} \right]\\ & = & \frac{d \mathbf{v}_1}{dt} \cdot \left[ \mathbf{v}_2 (t) \times \mathbf{v}_3(t)] + \frac{d \mathbf{v}_2}{dt} \cdot [ \mathbf{v}_3 (t) \times \mathbf{v}_1(t) ] + \frac{d \mathbf{v}_3}{dt} \cdot [\mathbf{v}_1 (t) \times \mathbf{v}_2 (t)\right] . \end{eqnarray*} At \( t = 0\), substitution from formula (\ref{equation3}) and the facts that \[ \mathbf{v}_1 \times \mathbf{v}_2 = \varepsilon \mathbf{v}_3, \quad \mathbf{v}_3 \times \mathbf{v}_1 = \varepsilon \mathbf{v}_2 \quad \mbox{and} \quad \mathbf{v}_2 \times \mathbf{v}_3 = \varepsilon \mathbf{v}_1, \] gives \begin{equation}\label{equation5} \left.\frac{d \mathcal{V}}{dt}\right|_{t = 0} = \varepsilon ^3 [\mathbf{D} \mathbf{F} (\mathbf{x}) \mathbf{i}] \cdot \mathbf{i} + \varepsilon ^3 [\mathbf{D}\mathbf{F}(\mathbf{x})\mathbf{j}]\cdot \mathbf{j} + \varepsilon^3 [\mathbf{D}\mathbf{F}(\mathbf{x})\mathbf{k}]\cdot\mathbf{k}.\tag{5} \end{equation} Since \( \mathbf{F} = F_1 \mathbf{i} + F_2 \mathbf{j}+F_3 \mathbf{k}\), we get \([\mathbf{D}\mathbf{F}(\mathbf{x})\mathbf{i}]\cdot \mathbf{i} = \partial F_1/\partial x\). Similarly, the second and third terms of equation (\ref{equation5}) are \(\varepsilon^3(\partial F_2/\partial y)\) and \( \varepsilon ^3 (\partial F _3 / \partial z \)). Substituting these facts into equation (\ref{equation5}) and dividing by \(\mathcal{V}(0) = \varepsilon ^3\) proves the theorem. \( \Box\)

example 2

Consider the vector field in the plane given by \({\bf V} (x,y) = x {\bf i}\). Relate the sign of the divergence of \({\bf V}\) with the rate of change of areas under the flow.

solution We think of \({\bf V}\) as the velocity field of a fluid in the plane. The vector field \({\bf V}\) points to the right for \(x > 0\) and to the left if \(x < 0\), as we see in Figure 14.53. The length of \({\bf V}\) gets shorter toward the origin. As the fluid moves, it expands (the area of the shaded rectangle increases), so we expect div \({\bf V} > 0\). Indeed, div \({\bf V} =1\).

Figure 14.53: This fluid is expanding.

example 3

The flow lines of the vector field \({\bf F} = x {\bf i}+ y {\bf j}\) are straight lines directed away from the origin (Figure 14.54).

Figure 14.54: The vector field \({\bf F}\hbox{(}x, y\hbox{)}=x{\bf i} + y\,{\bf j}\).

If these flow lines are those of a fluid, the fluid is expanding as it moves out from the origin, so div \({\bf F}\) should be positive. In fact, \[ \nabla \,{\cdot}\, {\bf F} = \frac{\partial}{\partial x} x+ \frac{\partial}{\partial y} y = 2 > 0. \]

example 4

Consider the vector field \({\bf F} = - x {\bf i} - y {\bf j}\). Here the flow lines point toward the origin instead of away from it (see Figure 14.55). Therefore, the fluid is compressing, so we expect \((\hbox{div }{\bf F}) < 0\). Calculating, we see that \[ \nabla \,{\cdot}\, {\bf F} = \frac{\partial}{\partial x} ( -x) + \frac{\partial}{\partial y} (-y) = -1 -1 = - 2 < 0. \]

Figure 14.55: The vector field \({\bf F}\hbox{(}x, y\hbox{)}= -x\,{\bf i} -y\,{\bf j}\).

example 5

As we saw in the last section, the flow lines of \({\bf F}=-y {\bf i}+ x {\bf j}\) are concentric circles about the origin, moving counterclockwise (see Figure 14.56). From this figure, it appears that the fluid is neither compressing nor expanding. This is confirmed by calculating \[ \nabla \,{\cdot}\, {\bf F} = \frac{\partial}{\partial x} (-y) + \frac{\partial}{\partial y} (x) = 0 + 0 =0. \]

Figure 14.56: The vector field \({\bf F}\hbox{(}x, y\hbox{)} = -y\, {\bf i} + x\,{\bf j}\) has zero divergence.

example 6

Some flow lines of \({\bf F} = x {\bf i} - y {\bf j}\) are shown in Figure 14.57. Here our intuition about expansion or compression is less clear. However, it is true that the shaded regions shown have the same area, and we calculate that \[ \nabla \,{\cdot}\, {\bf F}= \frac{\partial}{\partial x} x+ \frac{\partial}{\partial y} (-y)= 1+ (-1) =0. \]

Figure 14.57: The vector field \({\bf F}\hbox{(}x, y\hbox{)}=x{\bf i} - y\,{\bf j}\).

Curl of a Vector Field

If \({\bf F} = F_1 {\bf i}+ F_2 {\bf j} + F_3 {\bf k}\), the curl of \({\bf F}\) is the vector field \begin{eqnarray*} \hbox{curl } {\bf F} = \nabla \times {\bf F} & = & \left|\begin{array}{ccc} \displaystyle {\bf i } & {\bf j} & {\bf k} \\[2pt] \displaystyle \frac{\partial}{\partial x} & \displaystyle \frac{\partial}{\partial y} & \displaystyle \frac{\partial}{\partial z} \\[7pt] F_1 & F_2 & F_3 \end{array} \right| \\[6pt] & = & \bigg( \frac{\partial F_3}{\partial y} - \frac{\partial F_2}{\partial z} \bigg) {\bf i}+ \bigg( \frac{\partial F_1}{\partial z} - \frac{\partial F_3}{\partial x} \bigg) {\bf j}+ \bigg( \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y} \bigg) {\bf k.}\\[-18pt] \end{eqnarray*}

example 7

Let \({\bf F} (x,y,z) = x {\bf i}+ xy {\bf j} + {\bf k}\). Find \(\nabla \times {\bf F}.\)

solution We use the preceding formula: \[ \nabla \times {\bf F} = \left| \begin{array}{ccc} {\bf i} & {\bf j} & {\bf k} \\[2pt] \displaystyle \frac{\partial}{\partial x} & \displaystyle\frac{\partial}{\partial y} & \displaystyle\frac{\partial}{\partial z}\\[7pt] x & xy & 1 \end{array} \right|=(0-0){\bf i}-(0-0){\bf j}+(y-0){\bf k.} \]

Thus, \(\nabla \times {\bf F} = y {\bf k}.\)

example 8

Find the curl of \(xy {\bf i} - \sin z {\bf j} + {\bf k}\).

solution Letting \({\bf F} = xy {\bf i} -\sin z {\bf j} + {\bf k},\) \begin{eqnarray*} \nabla \times {\bf F} & = & \left| \begin{array}{ccc} {\bf i} & {\bf j} & {\bf k} \\[2pt] \displaystyle \frac{\partial}{\partial x} & \displaystyle \frac{\partial}{\partial y} & \displaystyle \frac{\partial}{\partial z} \\[7pt] xy& {-}{\sin z} & 1 \end{array} \right| \\[6pt] & = & \left| \begin{array}{cc} \displaystyle \frac{\partial}{\partial y} & \displaystyle \frac{\partial}{\partial z} \\[7pt] {-}{\sin z} & 1 \end{array} \right| {\bf i} - \left| \begin{array}{ccc} \displaystyle \frac{\partial}{\partial x} & \displaystyle \frac{\partial}{\partial z} \\[7pt] xy & 1 \end{array} \right| {\bf j} + \left| \begin{array}{ccc} \displaystyle \frac{\partial}{\partial x} & \displaystyle \frac{\partial}{\partial y} \\[7pt] xy & {-}{\sin z} \end{array} \right| {\bf k} \\[3pt] & = & \,\cos z {\bf i} - x {\bf k}. \\[-36pt] \end{eqnarray*}

example 9

Consider a solid rigid body \(B\) rotating about an axis \(L\). The rotational motion of the body can be described by a vector \(\boldsymbol{\omega}\) along the axis of rotation, the direction being chosen so that the body rotates about \(\boldsymbol{\omega}\), as in Figure 14.58. We call \(\boldsymbol{\omega}\) the angular velocity vector. The length \(\,\omega = \|\boldsymbol{\omega} \|\) is taken to be the angular speed of the body \(B\); that is, the speed of any point in \(B\) divided by its distance from the axis \(L\) of rotation. The motion of points in the rotating body is described by the vector field \({\bf v}\) whose value at each point is the velocity at that point. To find \({\bf v}\), let Q be any point in \(B\) and let \(\,\alpha\) be the distance from Q to \(L\).

Figure 14.58 shows that \(\,\alpha = \|{\bf r} \|\,\sin\,\theta\), where \({\bf r}\) is the vector whose initial point is the origin and whose terminal point is Q and \(\,\theta\) is the angle between \({\bf r}\) and the axis \(L\) of rotation. The tangential velocity \({\bf v}\) of Q is directed counterclockwise along the tangent to a circle parallel to the \(xy\) plane with radius \(\,\alpha\) and has magnitude \[ \|{\bf v} \| = \,\omega \,\alpha = \,\omega \|{\bf r} \|\,\sin \,\theta = \|\boldsymbol{\omega}\| \|{\bf r} \| \,\sin \,\theta. \]

The direction and magnitude of \({\bf v}\) imply that \({\bf v} = \boldsymbol{\omega} \times {\bf r}\). Selecting a coordinate system in which \(L\) is the \(z\) axis, we can write \(\boldsymbol{\omega} = \,\omega {\bf k}\) and \({\bf r} = x {\bf i} + y {\bf j}+ z {\bf k}\). Thus, \[ {\bf v} = \boldsymbol{\omega} \times {\bf r} = - \,\omega y {\bf i} + \,\omega x {\bf j}, \] and so \[ \hbox{curl } {\bf v} = \left| \begin{array}{ccc} {\bf i} & {\bf j} & {\bf k} \\[2pt] \displaystyle \frac{\partial}{\partial x} & \displaystyle \frac{\partial}{\partial y} & \displaystyle \frac{\partial}{\partial z} \\[7pt] -\omega y & \,\omega x & 0 \end{array} \right| = 2 \,\omega {\bf k} = 2\boldsymbol{\omega}. \]

Hence, for the rotation of a rigid body, the curl of the velocity vector field is a vector field whose value is the same at each point. It is directed along the axis of rotation with magnitude twice the angular speed.

Figure 14.58: The velocity \({\bf v}\) and angular velocity \(\boldsymbol{\omega}\) of a rotating body are related by \({\bf v} = \boldsymbol{\omega} \times {\bf r}\).

example 10

Verify that the vector field \[ {\bf V} (x,y,z) = \frac{y {\bf i} - x {\bf j}}{x^2 + y^2} \] is irrotational when \((x,y) \ne (0,0)\) (i.e., except where \({\bf V}\) is not defined).

solution The curl is \begin{eqnarray*} \nabla \times {\bf V} & = & \left| \begin{array}{ccc} {\bf i} & {\bf j} & {\bf k} \\[2pt] \displaystyle \frac{\partial}{\partial x} & \displaystyle \frac{\partial}{\partial y} & \displaystyle \frac{\partial}{\partial z} \\[12pt] \displaystyle\frac{y}{x^2 + y^2} & \displaystyle\frac{-x}{x^2 + y^2} & 0 \end{array} \right| \\[10pt] & = & 0 {\bf i} + 0 {\bf j} + \bigg[ \frac{\partial}{\partial x} \bigg( \frac{-x}{x^2 + y^2} \bigg) - \frac{\partial}{\partial y} \bigg( \frac{y}{x^2 + y^2} \bigg) \bigg] {\bf k} \\[15pt] & = & \bigg[ \frac{- (x^2 + y^2) + 2 x^2}{(x^2 + y^2)^2}+ \frac{- (x^2 + y^2) + 2 y^2}{(x^2 + y^2)^2} \bigg] {\bf k} = {\bf 0}. \\[-30pt] \end{eqnarray*}

Figure 14.59: Looking at a paddle wheel from above a moving fluid. The velocity field \({\bf V}\hbox{(}x, y, z\hbox{)} = \hbox{(}y\,{\bf i} - x\,{\bf j}\hbox{)}/\hbox{(}x^{2}+y^{\,2}\hbox{)}\) is irrotational; the paddle wheel does not rotate around its axis \(\boldsymbol{\omega}\).

Theorem 1 Curl of a Gradient

For any \(C^2\) function \(f\), \[ \nabla \times ( \nabla f) = {\bf 0}. \]

That is, the curl of any gradient is the zero vector.

proof

Because \(\nabla f = ( \partial f / \partial x , \partial f / \partial y, \partial f / \partial z)\) we have, by definition, \begin{eqnarray*} \nabla \times \nabla f & = & \left| \begin{array}{ccc} {\bf i} & {\bf j} & {\bf k} \\[2pt] \displaystyle \frac{\partial}{\partial x} & \displaystyle \frac{\partial}{\partial y} & \displaystyle \frac{\partial}{\partial z}\\[9pt] \displaystyle \frac{\partial f}{\partial x} & \displaystyle \frac{\partial f}{\partial y} & \displaystyle \frac{\partial f}{\partial z} \end{array} \right| \\[12pt] & = & \bigg( \frac{\partial^2 f}{\partial y \,\partial z} -\frac{\partial^2 f}{\partial z\, \partial y} \bigg) {\bf i} + \bigg( \frac{\partial^2 f}{\partial z \,\partial x} -\frac{\partial^2 f}{\partial x \,\partial z} \bigg) {\bf j}+ \bigg( \frac{\partial^2 f}{\partial x\, \partial y} -\frac{\partial^2 f}{\partial y \,\partial x} \bigg) {\bf k}. \end{eqnarray*}

Each component is zero because of the equality of mixed partial derivatives.

example 11

Let \({\bf V} (x,y,z) = y {\bf i} - x {\bf j}\). Show that \({\bf V}\) is not a gradient field.

solution If \({\bf V}\) were a gradient field, then it would satisfy curl \({\bf V} ={\bf 0}\) by Theorem 1. But \[ \hbox{curl } {\bf V} = \left| \begin{array}{ccc} {\bf i} & {\bf j} & {\bf k} \\[2pt] \displaystyle \frac{\partial }{\partial x} & \displaystyle \frac{\partial }{\partial y} & \displaystyle \frac{\partial }{\partial z} \\[7pt] y & - x & 0 \end{array} \right| = -2 {\bf k} \ne {\bf 0}, \] so \({\bf V}\) cannot be a gradient.

example 12

Find the scalar curl of \({\bf V} (x,y) = - y^2 {\bf i} + x {\bf j}\).

solution The curl is \[ \nabla \times {\bf V} = \left| \begin{array}{ccc} {\bf i} & {\bf j} & {\bf k} \\[2pt] \displaystyle \frac{\partial}{\partial x} &\displaystyle \frac{\partial}{\partial y} & \displaystyle \frac{\partial}{\partial z} \\[7pt] - y^2 & x & 0 \end{array} \right| = (1+2y)\, {\bf k}, \] so the scalar curl, which is the coefficient of \({\bf k}\), is \(1 + 2y\).

Theorem 2 Divergence of a Curl

For any \(C^2\) vector field \({\bf F},\) \[ \hbox{div } \hbox{curl } {\bf F} = \nabla \,{\cdot}\, ( \nabla \times {\bf F}) =0. \]

That is, the divergence of any curl is zero.

example 13

Show that the vector field \({\bf V} (x,y,z) = x {\bf i}+ y {\bf j} +z {\bf k}\) cannot be the curl of some vector field \({\bf F}\); that is, there is no \({\bf F}\) with \({\bf V} = \hbox{curl } {\bf F}\).

solution If this were so, then div \({\bf V}\) would be zero by Theorem 2. But \[ \hbox{div } {\bf V} = \frac{\partial x}{\partial x} + \frac{\partial y}{\partial y}+ \frac{\partial z}{\partial z} = 3 \ne 0, \] so \({\bf V}\) cannot be curl \({\bf F}\) for any \({\bf F}\).

example 14

Show that \(\nabla^2 f =0\) for \[ f (x,y,z) = \frac{1}{\sqrt{x^2 + y^2 + z^2}} = \frac{1}{r} \hbox{and} \qquad (x,y,z) \ne (0,0,0), \] where \({\bf r} = x {\bf i} + y {\bf j}+ z {\bf k}\) and \(r = \| {\bf r} \|\).

solution The first derivatives are \[ \frac{\partial f}{\partial x}= \frac{-x}{(x^2 + y^2 + z^2)^{3/2}},\quad \frac{\partial f}{\partial y}= \frac{-y}{(x^2 + y^2 + z^2)^{3/2}},\quad \frac{\partial f}{\partial z}= \frac{-z}{(x^2 + y^2 + z^2)^{3/2}}. \]

Computing the second derivatives, we find that \begin{eqnarray*} \frac{\partial^2 f}{\partial x^2} & = & \frac{3 x^2}{(x^2 + y^2 +z^2)^{5/2}} - \frac{1}{(x^2 + y^2 + z^2)^{3/2}} , \\[9pt] \frac{\partial^2 f}{\partial y^2} & = & \frac{3 y^2}{(x^2 + y^2 +z^2)^{5/2}} - \frac{1}{(x^2 + y^2 + z^2)^{3/2}} , \\[9pt] \frac{\partial^2 f}{\partial z^2} & = & \frac{3 z^2}{(x^2 + y^2 +z^2)^{5/2}} - \frac{1}{(x^2 + y^2 + z^2)^{3/2}} . \end{eqnarray*}

Thus, \begin{eqnarray*} \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2}+ \frac{\partial^2 f}{\partial z^2} & = & \frac{3 (x^2 + y^2 + z^2)}{(x^2 + y^2 + z^2)^{5/2}} - \frac{3}{(x^2 + y^2 + z^2)^{3/2}} \\[9pt] & = & \frac{3}{(x^2 + y^2 + z^2)^{3/2}} - \frac{3}{(x^2 + y^2 + z^2)^{3/2}}=0.\\[-30pt] \end{eqnarray*}

Basic Identities of Vector Analysis

1. \(\nabla ( f+ g) = \nabla f + \nabla g\)
2. \(\nabla (cf) =c \nabla f\), for a constant \(c\)
3. \(\nabla (fg) = f \nabla g + g \nabla f\)
4. \(\nabla (f/g) = (g \nabla f- f\nabla g) / g^2\), at points \({\bf x}\) where \(g ({\bf x}) \ne 0\)
5. \(\hbox{div } ({\bf F} + {\bf G}) = \hbox{div } {\bf F} + \hbox{div } {\bf G}\)
6. \(\hbox{curl } ({\bf F} + {\bf G}) = \hbox{curl } {\bf F} + \hbox{curl } {\bf G}\)
7. \(\hbox{div } ( f {\bf F}) = f \hbox{div } {\bf F} + {\bf F} \, {\cdot}\, \nabla f\)
8. \(\hbox{div } ( {\bf F} \times {\bf G}) = {\bf G}\,{\cdot}\, \hbox{curl } {\bf F} - {\bf F} \,{\cdot}\, \hbox{curl } {\bf G}\)
9. \(\hbox{div } \hbox{curl } {\bf F} =0\)
10. \(\hbox{curl } ( f {\bf F}) = f \hbox{curl } {\bf F} + \nabla f \times {\bf F}\)
11. \(\hbox{curl } \nabla f = {\bf 0}\)
12. \(\nabla^2 (fg) = f\nabla^2 g + g \nabla^2 f + 2 ( \nabla f \,{\cdot}\, \nabla g)\)
13. \(\hbox{div } ( \nabla f \times \nabla g) =0\)
14. \(\hbox{div } ( f\nabla g - g \nabla f) = f \nabla^2 g -g \nabla^2 f\)

example 15

Prove identity 7 in the preceding box.

solution The vector field \(f {\bf F}\) has components \(f F_i,\) for \(i =1,2,3\), and so \[ \hbox{div } ( f{\bf F}) = \frac{\partial}{\partial x} (f F_1) + \frac{\partial}{\partial y} ( f F_2) + \frac{\partial}{\partial z} ( f F_3). \]

However, \((\partial / \partial x)(f F_1) = f \partial F_1 / \partial x + F_1 \partial f / \partial x\) by the product rule, with similar expressions for the other terms. Therefore, \begin{eqnarray*} \hbox{div } ( f {\bf F}) & = & f \bigg( \frac{\partial F_1}{\partial x}+ \frac{\partial F_2}{\partial y} + \frac{\partial F_3}{\partial z} \bigg)+ F_1 \frac{\partial f}{\partial x} + F_2 \frac{\partial f}{\partial y} + F_3 \frac{\partial f}{\partial z} \\[9pt] & = & f ( \nabla \,{\cdot}\, {\bf F}) + {\bf F} \,{\cdot}\, \nabla f. \\[-30pt] \end{eqnarray*}

example 16

Show that for \({\bf r} \ne {\bf 0}, \nabla^2 ( 1/ r) =0\).

solution As in the case of the gravitational potential, \(\nabla (1/ r) = -{\bf r} / r^3\). In general, \(\nabla (r^n)=nr^{n-2}{\bf r}\) (see Exercise 38). By the identity \(\nabla \,{\cdot}\, ( f {\bf F}) = f\nabla \,{\cdot}\, {\bf F}+\nabla f \,{\cdot}\, {\bf F}\), we get \begin{eqnarray*} \nabla \,{\cdot}\, \bigg(\frac{{\bf r}}{r^3} \bigg) & = & \frac{1}{r^3} \nabla \,{\cdot}\, {\bf r} + {\bf r} \,{\cdot}\, \nabla \bigg( \frac{1}{r^3} \bigg) \\[1.5pt] & = & \frac{3}{r^3} + {\bf r} \,{\cdot}\, \bigg( \frac{-3 {\bf r}}{r^5}\bigg) = \frac{3}{r^3}-\frac{3}{r^3} =0.\\[-36pt] \end{eqnarray*}

Historical Note

Divergence and Curl

William Rowan Hamilton, in his investigation of quaternions (discussed in Section 11.3), introduced the del operator, defined formally as \[ \nabla=\frac{\partial}{\partial x}{\bf i} + \frac{\partial}{\partial y}\,{\bf j}+ \frac{\partial}{\partial z}{\bf k}. \]

Hamilton firmly believed in the significance of this operator. If \(f(x, y, z)\) is a scalar function on \({\mathbb R}^{3}\), then “multiplication” by \(\nabla\) gives the gradient of \(f\): \[ \nabla f=\frac{\partial f}{\partial x}{\bf i} + \frac{\partial f}{\partial y}\,{\bf j}+ \frac{\partial f}{\partial z}{\bf k}\hbox{,} \] which, of course, gives the direction of steepest ascent (see Section #). If \[ {\bf V}\hbox{(}x\hbox{,} y\hbox{,} z\hbox{)}=V_{1}\hbox{(}x\hbox{,} y\hbox{,} z\hbox{)} {\bf i} + V_{2}\hbox{(}x\hbox{,} y\hbox{,} z\hbox{)} {\bf j} + V_{3}\hbox{(}x\hbox{,} y\hbox{,} z\hbox{)}{\bf k} \] is a vector field, then the “quaternionic multiplication” of \(\nabla\) with \({\bf V}\) yields \[ \nabla {\bf V} = -\hbox{div } {\bf V} \,{+}\, \hbox{curl } {\bf V}. \]

Thus, what we now call the divergence of \({\bf V}\) is the negative of the scalar part of this product, and curl \({\bf V}\) is the vector part (c.f. the quaternion discussion in Section 11.3).

As far as we are aware, Hamilton never gave a physical interpretation of divergence and curl, but he surely believed that, as a consequence of his faith in them, they must have an important physical interpretation. His faith in his mathematical formalism was justified, but a physical explanation of divergence and curl had to wait for James Clerk Maxwell’s Treatise on Electricity and Magnetism. Here, Maxwell used both the divergence and the curl in his equations for the interaction of electric and magnetic fields (the Maxwell equations are discussed in Chapter 18).

Curiously, Maxwell referred to divergence as convergence and to curl as rotation, a term still used in scientific research. It was Josiah Gibbs (Figure ) who renamed convergence and rotation as the more familiar terms we use today—divergence and curl.

Figure 14.60: Josiah Willard Gibbs (1839–1903).

Maxwell gave a physical interpretation of the divergence using the Gauss divergence theorem, as we do in Section 18.4. His physical interpretation of the curl as a rotation was rather brief. Gibbs provided a more elementary interpretation of divergence, as we do in this section. In the spirit of Leibniz (who believed in infinitesimal quantities \({\it dx}\hbox{,} {\it dy}\hbox{,} {\it dz}\)), Gibbs imagined placing a small cube of dimensions \({\it dx}\) by \({\it dy}\) by \({\it dz}\) in a fluid. The faces of this cube have areas \({\it dx}\, {\it dy}\hbox{,}\) \({\it dy}\, {\it dz}\), and \({\it dx}\, {\it dz}\).

Historical Note

At this point, you may be interested to hear Gibbs through the words of his student E. B. Wilson:

Consider the amount of fluid which passes through those faces of the cube which are parallel to the \({\it yz}\) plane, i.e., perpendicular to the \({\it x}\) axis [see Figure 14.61].

The normal to the face whose \(x\) coordinate is the lesser, that is, the normal to the left-hand face of the cube is –\({\bf i}\). The flux of substance through this face is \[ -{\bf i} \,{\cdot}\, {\bf V} \hbox{(}x\hbox{,}\, y\hbox{,}\, z\hbox{)} \, {{\it dy}\, {\it dz}.} \]

Figure 14.61: Cube with faces parallel to the \({\it yz}\) plane.

The normal to the opposite face, the face whose \(x\) coordinate is greater by the amount \({\it dx}\), is \({+}{\bf i}\), and the flux through it is therefore \begin{eqnarray*} \displaystyle {\bf i} \,{\cdot}\, {\bf V} \hbox{(}x + {\it dx}\hbox{,}\, y\hbox{,}\, z\hbox{)} \, {{\it dy}\, {\it dz}} &=& {\bf i} \,{\cdot}\, \bigg[{\bf V} (x\hbox{,}\, y\hbox{,}\, z\hbox{)} + \frac{\partial {\bf V}}{\partial x}\,{\it dx}\bigg] \ {{\it dy} \,{\it dz}}\\[1.5pt] \displaystyle &=& {\bf i} \,{\cdot}\, {\bf V} \hbox{(}x\hbox{,}\, y\hbox{,}\, z\hbox{)} \, {{\it dy} \, {\it dz} } + {\bf i} \,{\cdot}\,\frac{\partial {\bf V}}{\partial x} \ {{\it dx}\, {\it dy}\, {\it dz}.} \end{eqnarray*}

The total flux outward from the cube through these two faces is therefore the algebraic sum of these quantities. This is simply \[ {\bf i} \,{\cdot}\,\frac{\partial {\bf V}}{\partial x}\, {{\it dx}\, {\it dy}\, {\it dz} } = \frac{\partial {V}_1}{\partial x}\, {{\it dx}\, {\it dy}\, {\it dz}.} \]

In like manner the fluxes through the other pairs of faces of the cube are \[ {\bf j} \,{{\cdot}}\,\frac{\partial {\bf V}}{\partial y}\, {{\it dx}\, {\it dy}\, {\it dz} }\qquad\hbox{and } \qquad{\bf k} \,{{\cdot}}\,\frac{\partial {\bf V}}{\partial z}\,{{\it dx}\, {\it dy}\, {\it dz}.} \]

The total flux out from the cube is therefore \[ \bigg({\bf i}\,{{\cdot}}\,\frac{\partial {\bf V}}{\partial x}+ {\bf j}\,{{\cdot}}\,\frac{\partial {\bf V}}{\partial y}+ {\bf k}\,{{\cdot}}\,\frac{\partial {\bf V}}{\partial z}\bigg)\, {{\it dx}\, {\it dy}\, {\it dz}.} \]

This is the net quantity of fluid that leaves the cube per unit time. The quotient of this by the volume \({{\it dx}\, {\it dy}\, {\it dz}}\) of the cube gives the rate of diminution of density. This is \[ \nabla \,{\cdot}\, {\bf V} = {\bf i}\,{{\cdot}}\,\frac{\partial {\bf V}}{\partial x} + {\bf j}\,{{\cdot}}\,\frac{\partial {\bf V}}{\partial y} + {\bf k}\,{{\cdot}}\,\frac{\partial {\bf V}}{\partial z} = \frac{\partial {V}_1}{\partial x} + \frac{\partial {V}_2}{\partial y} + \frac{\partial {V}_3}{\partial z}. \]

Because \(\nabla\,{\cdot}\,{\bf V}\) thus represents the diminution of density or the rate at which matter is leaving a point per unit volume per unit time, it is called the divergence. Maxwell employed the term convergence to denote the rate at which fluid approaches a point per unit volume per unit time. This is the negative of the divergence. In the case that the fluid is incompressible, as much matter must leave the cube as enters it. The total change of contents must therefore be zero. For this reason, the characteristic differential equation that any incompressible fluid must satisfy is \[ \nabla \,{\cdot}\, {\bf V} = \hbox{0}, \] where \({\bf V}\) is the velocity of the fluid. This equation is often known as the hydrodynamic equation. It is satisfied by any flow of water, since water is practically incompressible. The great importance of the equation for work in electricity is due to the fact that according to Maxwell’s hypothesis, electric displacement obeys the same laws as an incompressible fluid. If, then, \({\bf D}\) is the electric displacement, \[ \hbox{div } {\bf D} = \nabla \,{\cdot}\,{\bf D} = \hbox{0}. \]

Gibbs’ interpretation of curl was much like the one we gave in Example 9 for the rotation of a rigid body. Wilson remarks that an analysis of the meaning of curl for fluid motion was “rather difficult.” It remains a bit elusive, even today, as can be seen from our discussion following Example 9. We provide another interpretation in Chapter 18.