Divergence Identities

Let $\mathbf{F}(x, y, z) = P(x, y, z) \vec{i} + Q(x, y, z) \vec{j} + R(x, y, z) \vec{k}$ be a vector field on $\mathbb{R}^3$ and suppose that the necessary partial derivatives exist. Recall from The Divergence of a Vector Field page that the divergence of $\mathbf{F}$ can be computed with the following formula:

(1)

\begin{align} \quad \mathrm{div}( \mathbf{F}) = \nabla \cdot \mathbf{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z} \end{align}

Furthermore, from The Curl of a Vector Field page we saw that the curl of $\mathbf{F}$ can be computed with the following formula:

(2)

\begin{align} \quad \mathrm{curl} ( \mathbf{F} ) = \nabla \times \mathbf{F} = \left ( \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} \right ) \vec{i} + \left( \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x}\right ) \vec{j} + \left ( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right ) \vec{k} \end{align}

We will now look at a bunch of identities involving the divergence of a vector field. For all of the theorems above, we will assume the appropriate partial derivatives for the vector field $\mathbf{F} = P\vec{i} + Q\vec{j} + R \vec{k}$ and $w = f(x, y, z)$ exist and are continuous.

Theorem 1: Let $\mathbf{F} = P\vec{i} + Q\vec{j} + R \vec{k}$ and $\mathbf{G} = S\vec{i} + T\vec{j} + U\vec{k}$. If the partial derivatives of $P, Q, R, S, T, U$ exist then $\mathrm{div} (\mathbf{F} + \mathbf{G}) = \mathrm{div} (\mathbf{F}) + \mathrm{div} (\mathbf{G})$.

• Proof:

(3)

\begin{align} \quad \quad \mathrm{div} (\mathbf{F} + \mathbf{G}) = \nabla \cdot (\mathbf{F} + \mathbf{G}) = \nabla \cdot [(P + S) \vec{i} + (Q + T) \vec{j} + (R + U) \vec{k}] = \frac{\partial}{\partial x} (P + S) + \frac{\partial}{\partial y} (Q + T) + \frac{\partial}{\partial z} (R + U) \\ \quad \quad = \left ( \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \left ( \frac{\partial R}{\partial z} \right ) + \frac{\partial S}{\partial x} + \frac{\partial T}{\partial y} + \frac{\partial U}{\partial z} \right ) = \mathrm{div} (\mathbf{F}) + \mathrm{div} (\mathbf{G}) \quad \blacksquare \end{align}

Theorem 2: Let $\mathbf{F} = P\vec{i} + Q\vec{j} + R \vec{k}$ and $w = f(x, y, z)$ be a three variable real-valued function. If the partial derivatives of $P, Q, R, f$ exist then $\mathrm{div} (f \mathbf{F}) = f \mathrm{div} (\mathbf{F}) + \mathbf{F} \cdot \nabla f$.

• Proof:

(4)

\begin{align} \quad \quad \mathrm{div} (f \mathbf{F}) = \nabla \cdot (fP, fQ, fR) = \frac{\partial}{\partial x} (fP) + \frac{\partial}{\partial y} (fQ) + \frac{\partial}{\partial z} (fR) = \left ( f \frac{\partial P}{\partial x} + P \frac{\partial f}{\partial x} \right ) + \left ( f \frac{\partial Q}{\partial y} + Q \frac{\partial f}{\partial y} \right ) + \left ( f \frac{\partial R}{\partial z} + R \frac{\partial f}{\partial z} \right ) \\ \quad \quad = \left ( f \frac{\partial P}{\partial x} + f \frac{\partial Q}{\partial y} + f \frac{\partial R}{\partial z} \right ) + \left ( P \frac{\partial f}{\partial x} + Q \frac{\partial f}{\partial y} + R \frac{\partial f}{\partial z} \right ) = f \left ( \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z} \right ) + (P, Q, R) \cdot \left ( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right ) = f \mathrm{div} (\mathbf{F}) + \mathbf{F} \cdot \nabla f \quad \blacksquare \end{align}

Theorem 3: Let $\mathbf{F} = P\vec{i} + Q\vec{j} + R \vec{k}$ and $\mathbf{G} = S\vec{i} + T\vec{j} + U\vec{k}$. If the partial derivatives of $P, Q, R, S, T, U$ exist then $\mathrm{div} ( \mathbf{F} \times \mathbf{G}) = \mathbf{G} \cdot \mathrm{curl} ( \mathbf{F}) - \mathbf{F} \cdot \mathrm{curl} (\mathbf{G})$.

• Proof:

(5)

\begin{align} \quad \quad \mathrm{div} ( \mathbf{F} \times \mathbf{G} ) = \nabla \cdot (\mathbf{F} \times \mathbf{G}) = \nabla \cdot \begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\\ P & Q & R\\ S & T & U \end{vmatrix} = \nabla \cdot (QU - RT, RS - PU, PT - QS) \\ \quad \quad = \frac{\partial}{\partial x} (QU - RT) + \frac{\partial}{\partial y} (RS - PU) + \frac{\partial}{\partial z} (PT - QS) = \frac{\partial}{\partial x} (QU) - \frac{\partial}{\partial x} (RT) + \frac{\partial}{\partial y} (RS) - \frac{\partial}{\partial y} (PU) + \frac{\partial}{\partial z} (PT) - \frac{\partial}{\partial z} (QS) \\ \quad \quad = \left ( Q \frac{\partial U}{\partial x} + U \frac{\partial Q}{\partial x} \right ) - \left ( R \frac{\partial T}{\partial x} + T \frac{\partial R}{\partial x} \right ) + \left (R \frac{\partial S}{\partial y} + S \frac{\partial R}{\partial y} \right ) - \left ( P \frac{\partial U}{\partial y} + U \frac{\partial P}{\partial y} \right ) + \left (P \frac{\partial T}{\partial z} + T \frac{\partial P}{\partial z} \right ) - \left ( Q \frac{\partial S}{\partial z} + S \frac{\partial Q}{\partial z} \right ) \\ = \left [ S \left ( \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} \right ) + T \left ( \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x} \right ) + U \left ( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right ) \right ] + \left [ P \left ( \frac{\partial T}{\partial z} - \frac{\partial U}{\partial y} \right) + Q \left ( \frac{\partial U}{\partial x} - \frac{\partial S}{\partial z} \right ) + R \left ( \frac{\partial S}{\partial y} - \frac{\partial T}{\partial x} \right ) \right ] \\ = \left [ S \left ( \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} \right ) + T \left ( \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x} \right ) + U \left ( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right ) \right ] - \left [ P \left ( \frac{\partial U}{\partial y} - \frac{\partial T}{\partial z} \right) + Q \left ( \frac{\partial S}{\partial z} - \frac{\partial U}{\partial x} \right ) + R \left ( \frac{\partial T}{\partial x} - \frac{\partial S}{\partial y} \right ) \right ] \\ = \mathbf{G} \cdot \mathrm{curl} (\mathbf{F}) - \mathbf{F} \cdot \mathrm{curl} (\mathbf{G}) \quad \blacksquare \end{align}

Theorem 4: Let $w = f(x, y, z)$ and $w = g(x, y, z)$ be three variable real-valued functions. If the second partial derivatives of $f, g$ are continuous then $\mathrm{div} (\nabla f \times \nabla g) = 0$.

• Proof:

(6)

\begin{align} \quad \quad \mathrm{div} (\nabla f \times \nabla g) = \nabla \cdot (\nabla f \times \nabla g) = \nabla \cdot \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ \frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} & \frac{\partial f}{\partial z}\\ \frac{\partial g}{\partial x} & \frac{\partial g}{\partial y} & \frac{\partial g}{\partial z} \end{vmatrix} = \nabla \cdot \left ( \frac{\partial f}{\partial y} \frac{\partial g}{\partial z} - \frac{\partial f}{\partial z}\frac{\partial g}{\partial y} , \frac{\partial f}{\partial z} \frac{\partial g}{\partial x} - \frac{\partial f}{\partial x} \frac{\partial g}{\partial z} , \frac{\partial f}{\partial x} \frac{\partial g}{\partial y} - \frac{\partial f}{\partial y} \frac{\partial g}{\partial x} \right) \\ \quad \quad = \frac{\partial}{\partial x} \left ( \frac{\partial f}{\partial y} \frac{\partial g}{\partial z} - \frac{\partial f}{\partial z}\frac{\partial g}{\partial y} \right ) + \frac{\partial}{\partial y} \left ( \frac{\partial f}{\partial z} \frac{\partial g}{\partial x} - \frac{\partial f}{\partial x} \frac{\partial g}{\partial z} \right ) + \frac{\partial}{\partial z} \left ( \frac{\partial f}{\partial x} \frac{\partial g}{\partial y} - \frac{\partial f}{\partial y} \frac{\partial g}{\partial x} \right ) \\ = \left[\left ( \frac{\partial f}{\partial y} \frac{\partial^2 g}{\partial x \partial z} + \frac{\partial g}{\partial z} \frac{\partial^2 f}{\partial x \partial y}\right ) - \left ( \frac{\partial f}{\partial z}\frac{\partial^2 g}{\partial x \partial y} + \frac{\partial g}{\partial y} \frac{\partial^2 f}{\partial x \partial z} \right ) \right ] + \left [ \left ( \frac{\partial f}{\partial z} \frac{\partial^2 g}{\partial y \partial x} + \frac{\partial g}{\partial x} \frac{\partial^2 f}{\partial y \partial z} \right ) - \left ( \frac{\partial f}{\partial x} \frac{\partial^2 g}{\partial y \partial z} + \frac{\partial g}{\partial z} \frac{\partial^2 f}{\partial y \partial x} \right ) \right ] \\ + \left [ \left ( \frac{\partial f}{\partial x} \frac{\partial^2 g}{\partial z \partial y} + \frac{\partial g}{\partial y}\frac{\partial^2 f}{\partial z \partial x} \right ) - \left ( \frac{\partial f}{\partial y} \frac{\partial^2 g}{\partial z \partial x} + \frac{\partial g}{\partial x} \frac{\partial^2 f}{\partial z \partial y} \right ) \right ] \end{align}

• By Clairaut's Theorem on Higher Order Partial Derivatives, the mixed second partial derivatives of $f$ and $g$ are equal, and from above, we can cancel each term out to get that $\mathrm{div} (\nabla f \times \nabla g) = 0$ as desired. $\blacksquare$