Curl Identities

# Curl 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 curl 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 $\mathrm{curl} (\mathbf{F} + \mathbf{G}) = \mathrm{curl} (\mathbf{F}) + \mathrm{curl} (\mathbf{G})$.
• Proof:
(3)
\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}
 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 $\mathrm{curl} (f \mathbf{F}) = f \mathrm{curl} (\mathbf{F}) + \nabla f \times \mathbf{F}$.
• Proof:
(4)
\begin{align} \quad \quad \mathrm{curl} (f \mathbf{F}) = \nabla \times (f \mathbf{F}) = \nabla \times (fP, fQ, fR) = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} &\frac{\partial}{\partial z} \\ fP & fQ & fR \end{vmatrix} = \left ( \frac{\partial}{\partial y} (fR) - \frac{\partial}{\partial z} (fQ)\right ) \vec{i} + \left ( \frac{\partial}{\partial z} (fP) - \frac{\partial}{\partial x} (fR) \right ) \vec{j} + \left ( \frac{\partial}{\partial x} (fQ) - \frac{\partial}{\partial y} (fP) \right ) \vec{k} \\ = \left [ \left ( f \frac{\partial R}{\partial y} + R\frac{\partial f}{\partial y} \right ) - \left ( f \frac{\partial Q}{\partial z} + Q \frac{\partial f}{\partial z} \right ) \right ] \vec{i} + \left [ \left ( f \frac{\partial P}{\partial z} + P \frac{\partial f}{\partial z} \right ) - \left ( f \frac{\partial R}{\partial x} + R \frac{\partial f}{\partial x} \right ) \right ] \vec{j} + \left [ \left (f \frac{\partial Q}{\partial x} + Q \frac{\partial f}{\partial x} \right ) - \left ( f \frac{\partial P}{\partial y} + P \frac{\partial f}{\partial y} \right ) \right ] \vec{k} \\ = f \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 ( R \frac{\partial f}{\partial y} - Q \frac{\partial f}{\partial z} \right ) \vec{i} + \left ( P \frac{\partial f}{\partial z} - R \frac{\partial f}{\partial x} \right ) \vec{j} + \left ( Q \frac{\partial f}{\partial x} - P \frac{\partial f}{\partial y} \right ) \vec{j} \right ] \\ = f \mathrm{curl} (\mathbf{F}) + \nabla f \times \mathbf{F} \quad \blacksquare \end{align}
 Theorem 3: Let $\mathbf{F} = P\vec{i} + Q\vec{j} + R \vec{k}$. If the second partial derivatives of $P, Q, R$ are continuous then $\mathrm{curl} (\mathrm{curl} ( \mathbf{F} )) = \nabla (\mathrm{div} ( \mathbf{F})) - \nabla^2 \mathbf{F}$ where $\nabla^2 \mathbf{F} = \nabla^2 P \vec{i} + \nabla^2 Q \vec{j} + \nabla^2 R \vec{k}$
• Proof:
(5)
\begin{align} \quad \quad \mathrm{curl} (\mathrm{curl} ( \mathbf{F} )) = \nabla \times (\nabla \times \mathbf{F}) = \nabla \times \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 ] = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} &\frac{\partial}{\partial z} \\ \left ( \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} \right ) & \left( \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x}\right ) & \left ( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right ) \end{vmatrix} \\ \quad \quad = \left [ \frac{\partial}{\partial y} \left ( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right ) - \frac{\partial}{\partial z} \left( \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x}\right ) \right ] \vec{i} +\left [ \frac{\partial}{\partial z} \left ( \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} \right ) - \frac{\partial}{\partial x} \left ( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right ) \right ] \vec{j} + \left [ \frac{\partial}{\partial x}\left( \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x}\right ) - \frac{\partial}{\partial y}\left ( \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} \right ) \right ] \vec{k} \\ \quad \quad = \left [ \left ( \frac{\partial^2 Q}{\partial y \partial x} - \frac{\partial^2 P}{\partial y^2} \right ) - \left ( \frac{\partial^2 P}{\partial z^2} + \frac{\partial^2 R}{\partial z \partial x}\right ) \right ] \vec{i} + \left [ \left ( \frac{\partial^2 R}{\partial z \partial y} - \frac{\partial^2 Q}{\partial z^2} \right ) - \left( \frac{\partial^2 Q}{\partial x^2} - \frac{\partial^2 P}{\partial x \partial y} \right ) \right ] \vec{j} + \left [ \left ( \frac{\partial^2 P}{\partial x \partial z} - \frac{\partial^2 R}{\partial x^2}\right ) - \left ( \frac{\partial^2 R}{\partial y^2} - \frac{\partial^2 Q}{\partial y \partial z} \right )\right ] \vec{k} \end{align}
• Now we will compute $\nabla (\mathrm{div} (\mathbf{F})) - \nabla^2 \mathbf{F}$.
(6)
\begin{align} \quad \quad \nabla (\mathrm{div} (\mathbf{F})) - \nabla^2 \mathbf{F} = \nabla \left ( \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z}\right ) - \left [ \left ( \frac{\partial^2 P}{\partial x^2} + \frac{\partial^2 P}{\partial y^2} + \frac{\partial^2 P}{\partial z^2} \right ) \vec{i} + \left(\frac{\partial^2 Q}{\partial x^2} + \frac{\partial^2 Q}{\partial y^2} + \frac{\partial^2 Q}{\partial z^2} \right ) \vec{j} + \left ( \frac{\partial^2 R}{\partial x^2} + \frac{\partial^2 R}{\partial y^2} + \frac{\partial^2 R}{\partial z^2} \right )\vec{k} \right ] \\ = \left [ \left ( \frac{\partial^2 P}{\partial x^2} + \frac{\partial^2 Q}{\partial x \partial y} + \frac{\partial^2 R}{\partial x \partial z} \right ) \vec{i} + \left ( \frac{\partial^2 P}{\partial y \partial x} + \frac{\partial^2 Q}{\partial y^2} + \frac{\partial^2 R}{\partial y \partial z} \right ) \vec{j} + \left( \frac{\partial^2 P}{\partial z \partial x} + \frac{\partial^2 Q}{\partial z \partial y} + \frac{\partial^2 R}{\partial z^2} \right ) \vec{k} \right ] \\ - \left [ \left ( \frac{\partial^2 P}{\partial x^2} + \frac{\partial^2 P}{\partial y^2} + \frac{\partial^2 P}{\partial z^2} \right ) \vec{i} + \left(\frac{\partial^2 Q}{\partial x^2} + \frac{\partial^2 Q}{\partial y^2} + \frac{\partial^2 Q}{\partial z^2} \right ) \vec{j} + \left ( \frac{\partial^2 R}{\partial x^2} + \frac{\partial^2 R}{\partial y^2} + \frac{\partial^2 R}{\partial z^2} \right )\vec{k} \right ] \\ = \left (\frac{\partial^2 Q}{\partial x \partial y} + \frac{\partial^2 R}{\partial x \partial z} - \frac{\partial^2 P}{\partial y^2} - \frac{\partial^2 P}{\partial z^2} \right ) \vec{i} + \left ( \frac{\partial^2 P}{\partial y \partial x} + \frac{\partial^2 R}{\partial y \partial z} - \frac{\partial^2 Q}{\partial x^2} - \frac{\partial^2 Q}{\partial z^2} \right) \vec{j} + \left ( \frac{\partial^2 P}{\partial z \partial x} + \frac{\partial^2 Q}{\partial z \partial y} - \frac{\partial^2 R}{\partial x^2} - \frac{\partial^2 R}{\partial y^2} \right ) \vec{k} \end{align}
• By Clairaut's Theorem on Higher Order Partial Derivatives we have that the second order mixed partial derivatives of $P$, $Q$, and $R$ are equal, and thus, with some rearranging of the last equation above, we can see that indeed $\mathrm{curl} (\mathrm{curl} ( \mathbf{F} )) = \nabla (\mathrm{div} ( \mathbf{F})) - \nabla^2 \mathbf{F}$. $\blacksquare$