Differential Operators

Certain differential operations may be performed on scalar and vector fields and have wide range of applications. Central to all these differential operations is the following vector operator.

Definition. The vector operator del/nabla, denoted by $\nabla$, in Cartesian coordinate systems is defined by

\[\nabla \cong \mathbf{e}_j{\partial \over \partial x_j}\]

Gradient

We start with differentiating a scalar field and assume we are working with Cartesian coordinates.

Definition. The gradient of a scalar field $\phi: \mathbb{R}^m \to \mathbb{R}$ is a vector field defined by

\[\text{grad} \,\phi = \nabla \phi \cong \mathbf{e}_j { \partial \phi \over \partial x_j } = \left({\partial \phi \over \partial x_1}, {\partial \phi \over \partial x_2}, ..., {\partial \phi \over \partial x_m} \right)\]

Proposition. The directional derivative of $\phi$ in the direction of $\mathbf{u}$ is given by

\[D_\mathbf{u} \,\phi = \nabla \phi \cdot \mathbf{u}\]

Proof.

The directional derivative is given by

\[D_\mathbf{u} \,\phi = {\partial \phi \over \partial x_j} u_j = \mathbf{e_j} {\partial \phi \over \partial x_j} \mathbf{e_i} \,u_i = \nabla \phi \cdot \mathbf{u}\]

Corollary. $\nabla \phi$ lies in the direction of the fastest rate of increase in $\phi$. The largest rate of decrease of $\phi$ is in the direction of $-\nabla \phi$.

Proposition. Suppose that $\mathbf{r}(u)$ is a curve and $\phi: \mathbb{R}^m \to \mathbb{R}$ is a scalar field, then the derivative of $\phi(\mathbf{r}(u))$ along the curve is given by

\[{d \over du} \phi(\mathbf{r}(u)) = \nabla \phi \cdot {d \mathbf{r} \over du}\]

If the parameter is the arc length $s$ then $d\phi/ds = \nabla \phi \cdot \mathbf{t}$.

Proof.

By chain rule,

\[{d \over du} \phi(\mathbf{r}(u)) = {\partial \phi \over \partial x_j} {dx_j \over du}\]

By definition,

\[{d\mathbf{r} \over du} = \mathbf{e}_j {dx_j \over du}\]

Hence,

\[\nabla \phi \cdot {d\mathbf{r} \over du} = \mathbf{e}_j {\partial f \over \partial x_j} \mathbf{e}_i {dx_i \over du} = \delta_{ij} {\partial f \over \partial x_j} {dx_i \over du} = {\partial f \over \partial x_j} {dx_j \over du} = {d \over du} \phi(\mathbf{r}(u))\]

Proposition. The differential $d\phi$ along the curve $\mathbf{r}(u)$ is given by

\[d\phi = \nabla \phi \cdot d\mathbf{r}\]

Proof.

\[\nabla \phi \cdot d\mathbf{r} = \mathbf{e}_j {\partial f \over \partial x_j} \mathbf{e}_i \,dx_i = d\phi\]

Property. [Linearity] Suppose that $\phi$ and $\psi$ are scalar fields and $\lambda, \mu \in \mathbb{R}$. Then

\[\nabla (\lambda \phi + \mu \psi) = \lambda(\nabla \phi) + \mu(\nabla \psi)\]

Proof.

\[\nabla (\lambda \phi + \mu \psi) = \mathbf{e}_j \left[{\partial \over \partial x_j} (\lambda \phi + \mu \psi)\right] = \lambda \left(\mathbf{e}_j {\partial \phi \over \partial x_j}\right) + \mu \left(\mathbf{e}_j {\partial \psi \over \partial x_j}\right) = \lambda(\nabla \phi) + \mu(\nabla \psi)\]

Divergence

By treating $\nabla$ as an vector and an operator that is waiting for a function to come along and be differentiated, we can develop how it interacts with vector fields.

Definition. The divergence of a vector field $\mathbf{F}: \mathbb{R}^n \to \mathbb{R}^n$ is a scalar field defined by

\[\text{div} \,\phi = \nabla \cdot \mathbf{F} = (\mathbf{e}_i {\partial \over \partial x_i}) \cdot (\mathbf{e}_j \,F_j) = {\partial F_j \over \partial x_j}\]

For a vector function $\mathbf{F}: \mathbb{R}^3 \to \mathbb{R}^3 = (F_1, F_2, F_3)$,

\[\nabla \cdot \mathbf{F} = {\partial F_1 \over \partial x} + {\partial F_2 \over \partial y} + {\partial F_3 \over \partial z}\]

Definition. A vector field $\mathbf{F}$ is solenoidal if $\nabla \cdot \mathbf{F} = 0$.

Proposition. The divergence of a vector field is the trace of its Jacobian matrix.

Property. [Linearity] Suppose that $\mathbf{F}$ and $\mathbf{G}$ are vector fields and $\lambda, \mu \in \mathbb{R}$. Then

\[\nabla \cdot (\lambda \mathbf{F} + \mu \mathbf{G}) = \lambda(\nabla \cdot \mathbf{F}) + \mu(\nabla \cdot \mathbf{G})\]

Proof.

\[\nabla \cdot (\lambda \mathbf{F} + \mu \mathbf{G}) = {\partial \over \partial x_j} (\lambda F_j + \mu G_j) = \lambda {\partial F_j \over \partial x_j} + \mu {\partial G_j \over \partial x_j} = \lambda(\nabla \cdot \mathbf{F}) + \mu(\nabla \cdot \mathbf{G})\]

Curl

Definition. The curl of a vector field $\mathbf{F}: \mathbb{R}^3 \to \mathbb{R}^3$ is a vector field defined by

\[\begin{align*} \text{curl} \,\mathbf{F} = \nabla \times \mathbf{F} &= (\mathbf{i} \partial_x + \mathbf{j} \partial_y + \mathbf{j} \partial_z) \times (\mathbf{i} F_x + \mathbf{j} F_y + \mathbf{j} F_z) \\ &= \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \partial_x & \partial_y & \partial_k \\ F_x & F_y & F_k \\ \end{vmatrix} \\ &= (\partial_y F_z - \partial_z F_y, \partial_z F_x - \partial_x F_z, \partial_x F_y - \partial_y F_x) \\ &= \varepsilon_{ijk} \,\mathbf{e}_i \,{\partial F_k \over \partial x_j} \end{align*}\]

Proposition. Let $\mathsf{T}$ be the Jacobian matrix of $\mathbf{F}$. Then

\[\nabla \times \mathbf{F} = \varepsilon_{ijk} \,\mathbf{e}_i T_{kj}\]

Definition. A vector field $\mathbf{F}$ is irrotational if $\nabla \times \mathbf{F} = 0$.

Property. [Linearity] Suppose that $\mathbf{F}$ and $\mathbf{G}$ are vector fields and $\lambda, \mu \in \mathbb{R}$. Then

\[\nabla \times (\lambda \mathbf{F} + \mu \mathbf{G}) = \lambda(\nabla \times \mathbf{F}) + \mu(\nabla \times \mathbf{G})\]

Proof.

\[\begin{align*} \nabla \times (\lambda \mathbf{F} + \mu \mathbf{G}) &= \varepsilon_{ijk} \,\mathbf{e}_i \,{\partial \over \partial x_j}(\lambda F_k + \mu G_k) \\ &= \lambda \left(\varepsilon_{ijk} \,\mathbf{e}_i \,{\partial F_k \over \partial x_j}\right) + \mu \left(\varepsilon_{ijk} \,\mathbf{e}_i \,{\partial G_k \over \partial x_j}\right) \\ &= \lambda(\nabla \times \mathbf{F}) + \mu(\nabla \times \mathbf{G}) \end{align*}\]

Vector Differential Identities

Definition. Suppose $\mathbf{F}$ is a vector field, $\mathbf{F} \cdot \nabla$ is the differential operator

\[\mathbf{F} \cdot \nabla \cong F_j {\partial \over \partial x_j}\]

For a scalar field $\phi: \mathbb{R}^3 \to \mathbb{R}$,

\[(\mathbf{F} \cdot \nabla)\phi = F_j {\partial \phi \over \partial x_j} = F_1 {\partial \phi \over \partial x} + F_2 {\partial \phi \over \partial y} + F_3 {\partial \phi \over \partial z} = \mathbf{F} \cdot (\nabla \phi)\]

For a vector field $\mathbf{G}: \mathbb{R}^3 \to \mathbb{R}^3$,

\[\begin{align*} (\mathbf{F} \cdot \nabla)\mathbf{G} = \mathbf{e}_i \,F_j {\partial G_i \over \partial x_j} &= \left(F_1 {\partial G_1 \over \partial x} + F_2 {\partial G_1 \over \partial y}+ F_3 {\partial G_1 \over \partial z} \right) \,\mathbf{i} \\ &\qquad + \left(F_1 {\partial G_2 \over \partial x} + F_2 {\partial G_2 \over \partial y}+ F_3 {\partial G_2 \over \partial z} \right) \,\mathbf{j} \\ &\qquad + \left(F_1 {\partial G_3 \over \partial x} + F_2 {\partial G_3 \over \partial y}+ F_3 {\partial G_3 \over \partial z} \right) \,\mathbf{k} \end{align*}\]

Proposition. [Leibniz Property] Suppose that $\phi$ and $\psi$ are scalar fields and $\mathbf{F}$ is a vector field. Then

\[\begin{align*} \nabla (\phi \psi) &= (\nabla \phi)\psi + \phi (\nabla \psi) \\ \nabla \cdot (\phi \mathbf{F}) &= (\nabla \phi) \cdot \mathbf{F} + \phi(\nabla \cdot \mathbf{F}) \\ \nabla \times (\phi \mathbf{F}) &= (\nabla \phi) \times \mathbf{F} + \phi(\nabla \times \mathbf{F}) \\ \end{align*}\]

Proposition. Suppose that $\mathbf{F}$ and $\mathbf{G}$ are vector fields. Then

\[\begin{align*} \nabla \cdot (\mathbf{F} \times \mathbf{G}) &= (\nabla \times \mathbf{F}) \cdot \mathbf{G} - \mathbf{F} \cdot (\nabla \times \mathbf{G}) \\ \nabla \times (\mathbf{F} \times \mathbf{G}) &= (\nabla \cdot \mathbf{G}) \mathbf{F} - (\nabla \cdot \mathbf{F}) \mathbf{G} + (\mathbf{G} \cdot \nabla) \mathbf{F} - (\mathbf{F} \cdot \nabla) \mathbf{G} \\ \nabla (\mathbf{F} \cdot \mathbf{G}) &= \mathbf{F} \times (\nabla \times \mathbf{G}) + \mathbf{G} \times (\nabla \times \mathbf{F}) + (\mathbf{F} \cdot \nabla) \mathbf{G} + (\mathbf{G} \cdot \nabla) \mathbf{F} \\ \end{align*}\]

Second Order Differential Operators

There are a number of possible ways to combining the differential operators.

Divergence of Gradient

Consider the divergence of the gradient of a scalar field $\phi: \mathbb{R}^3 \to \mathbb{R}$, by definition

\[\nabla \cdot (\nabla \phi) = {\partial \over \partial x_j} \left({\partial \phi \over \partial x_j }\right) = {\partial^2 \phi \over \partial x^2} + {\partial^2 \phi \over \partial y^2} + {\partial^2 \phi \over \partial z^2}\]

Definition. The Laplacian operator $\nabla^2$ is defined by

\[\nabla^2 = \nabla \cdot \nabla = { \partial^2 \over \partial x_j \partial x_j }\]

Divergence of Curl

Proposition. The divergence of the curl of vector field $\mathbf{F}$ with equal mixed partial derivatives is always zero, i.e.

\[\nabla \cdot (\nabla \times \mathbf{F}) = 0\]

Proof.

\[\nabla \cdot (\nabla \times \mathbf{F}) = { \partial \over \partial x_i } \left( \varepsilon_{ijk} {\partial F_k \over \partial x_j} \right) = \varepsilon_{ijk} {\partial f_k \over \partial x_i \partial x_j} = 0\]

since $\partial_i \partial_j F_k = \partial_j \partial_j F_k$.

Proposition. Conversely, suppose that a vector field $\mathbf{v}$ is solenoidal, i.e. $\nabla \cdot \mathbf{v} = 0$, then there exists a non-unique vector potential $\mathbf{A}$ such that $\nabla \times \mathbf{A} = \mathbf{v}$.

It is relatively straightforward to construct a vector potential. Let $\mathbf{v} = (v_1, v_2, v_3)$ to be defined throughout $\mathbb{R}^3$, then a possible vector potential is

\[\mathbf{A}(\mathbf{x}) = \left( \int_{z_0}^z v_2(x, y, z') \,dz', \int_{x_0}^x v_3(x', y, z_0) \,dx' - \int_{z_0}^z v_1(x, y, z') \,dz', 0 \right)\]

where $\mathbf{x}_0 = (x_0, y_0, z_0)$ is a constant.

Curl of Gradient

Proposition. The curl of the gradient of a scalar field $\phi$ with equal mixed partial derivatives is always zero, i.e.

\[\nabla \times (\nabla \phi) = \mathbf{0}\]

Proof.

\[[\nabla \times (\nabla \phi)]_i = \varepsilon_{ijk} {\partial \over \partial x_j} \left( { \partial \phi \over \partial x_k } \right) = \varepsilon_{ijk} {\partial^2 \phi \over \partial x_j \partial x_k} = 0\]

since $\partial_j \partial_k \phi = \partial_k \partial_j \phi$.

The vector potential defined above is not unique since if $\nabla \times \mathbf{A} = \mathbf{v}$, then

\[\nabla \times (\mathbf{A} + \nabla \phi) = \nabla \times \mathbf{A} = \mathbf{v}\]

for all scalar functions $\phi$.

Curl of Curl

Consider the curl of the curl of a vector field $\mathbf{F}$ with equal mixed derivatives, then

\[\begin{align*} [\nabla \times (\nabla \times \mathbf{F})]_i &= \varepsilon_{ijk} {\partial \over \partial x_j} \left( \varepsilon_{kpq} { \partial F_q \over \partial x_p } \right) \\ &= (\delta_{ip}\delta_{jq} - \delta_{iq}\delta_{jp}) {\partial^2 F_q \over \partial x_j \partial x_p} \\ &= {\partial^2 F_j \over \partial x_j \partial x_i } - {\partial^2 F_i \over \partial x_j \partial x_j } \\ &= [\nabla(\nabla \cdot \mathbf{F}) - \nabla^2 \mathbf{F}]_i \end{align*}\]

Definition. The Laplacian operator $\nabla^2$ acting on a vector function $\mathbf{F}$ is defined by

\[\nabla^2 \mathbf{F} = \nabla(\nabla \cdot \mathbf{F}) - \nabla \times (\nabla \times \mathbf{F})\]

References