Curl and Divergence Identities

# Curl and 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 relating the concepts of curl, divergence, and gradient. All of the proofs given use the definitions given above and are relatively straightforward. 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 $\mathbf{G} = S\vec{i} + T\vec{j} + U\vec{k}$. Then $\mathrm{curl} (\mathbf{F} + \mathbf{G}) = \mathrm{curl} (\mathbf{F}) + \mathrm{curl} (\mathbf{G})$.
• Proof:
(4)
\begin{align} \quad \quad \mathrm{curl} (\mathbf{F} + \mathbf{G}) = \nabla \times (\mathbf{F} + \mathbf{G}) = \nabla \times [(P + S) \vec{i} + (Q + T) \vec{j} + (R + U) \vec{k}] = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z}\\ P + S & Q + T & R + U \end{vmatrix} \\ \quad = \left ( \frac{\partial}{\partial y} (R + U) - \frac{\partial}{\partial z} (Q + T) \right ) \vec{i} + \left ( \frac{\partial}{\partial z} (P + S) - \frac{\partial}{\partial x} (R + U)\right ) \vec{j} + \left ( \frac{\partial}{\partial x} (Q + T) - \frac{\partial}{\partial y} (P + S) \right ) \\ \quad \: \quad = \left [ \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} \right ] + \left [ \left ( \frac{\partial U}{\partial y} - \frac{\partial T}{\partial z} \right ) \vec{i} + \left( \frac{\partial S}{\partial z} - \frac{\partial U}{\partial x}\right ) \vec{j} + \left ( \frac{\partial T}{\partial x} - \frac{\partial S}{\partial y} \right ) \vec{k} \right ] \\ \quad = \mathrm{curl} (\mathbf{F}) + \mathrm{curl} ( \mathbf{G} ) \quad \blacksquare \end{align}