Properties of The Divergence and Curl of a Vector Field

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 some important theorems regarding the curl and divergence of a vector field $\mathbf{F}$ and the gradient $\nabla f$ of a three variable real-valued function $w = f(x, y, z)$.

Theorem 1: Let $w = f(x, y, z)$ be a three variable real-valued function with continuous second partial derivatives. Then $\mathrm{curl} (\nabla f) = \vec{0}$.

• Proof: Suppose that $f$ has continuous second partial derivatives. Using the definition of curl with respect to the vector field $\nabla f(x, y, z) = \frac{\partial f}{\partial x} \vec{i} + \frac{\partial f}{\partial y} \vec{y} + \frac{\partial f}{\partial z} \vec{k}$ and we thus have that:

(3)

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

• Since $f$ has continuous second partial derivatives, it follows by Clairaut's Theorem on Higher Order Partial Derivatives that the mixed second partial derivatives of $f$ equal each other, and thus $\mathrm{curl} (\nabla f) = \vec{0}$. $\blacksquare$

Theorem 2: 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 let $P$, $Q$, and $R$ have continuous second partial derivatives. Then $\mathrm{div} ( \mathrm{curl} ( \mathbf{F})) = 0$.

• Proof: Suppose that $P$, $Q$, and $R$ have continuous second partial derivatives. Then:

(4)

\begin{align} \quad \quad \mathrm{div} ( \mathrm{curl} ( \mathbf{F} )) = \nabla \cdot \mathrm{curl} (\mathbf{F}) = \nabla \cdot (\nabla \times \mathbf{F}) = \frac{\partial}{\partial x} \left ( \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} \right ) + \frac{\partial}{\partial y} \left( \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x}\right ) + \frac{\partial}{\partial z} \left ( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right ) \\ = \frac{\partial^2 R}{\partial x \partial y} - \frac{\partial^2 Q}{\partial x \partial z} + \frac{\partial^2 P}{\partial y \partial z} - \frac{\partial^2 R}{\partial y \partial x} + \frac{\partial^2 Q}{\partial z \partial x} - \frac{\partial^2 P}{\partial z \partial y} = \left ( \frac{\partial^2 R}{\partial x \partial y} - \frac{\partial^2 R}{\partial y \partial x} \right ) + \left ( \frac{\partial^2 Q}{\partial z \partial x} - \frac{\partial^2 Q}{\partial x \partial z} \right ) + \left ( \frac{\partial^2 P}{\partial y \partial z} - \frac{\partial^2 P}{\partial z \partial y} \right ) \end{align}

• Since $P$, $Q$, and $R$ all have continuous second partial derivatives, it follows from Clairaut's Theorem that the mixed second partial derivatives of $f$ equal each other, and thus $\mathrm{div} (\mathrm{curl} ( \mathbf{F} )) = 0$. $\blacksquare$