We take a fresh look at the vector differential operators grad, div and curl. There are many vector identities relating these. In particular, there are two combinations that always yield zero results:

$\displaystyle \begin{array}{rcl} \mathbf{curl}\ \mathbf{grad}\ \chi &\equiv& 0\,, \quad \mbox{for all scalar functions\ }\chi \\ \mathrm{div}\ \mathbf{curl}\ \boldsymbol{\psi} &\equiv& 0\,, \quad \mbox{for all vector functions\ }\boldsymbol{\psi} \end{array}$

Question: Is there a connection between these identities?

A Hilbert space ${H}$ is a function space that is complete and has an inner product: for ${f}$ and ${g}$ in ${H}$, there is a mapping from ${H\times H}$ to the real numbers: ${\langle f , g \rangle \in \mathbb{R}}$. We consider a linear operator ${\mathsf{A}}$ mapping one Hilbert space to another, ${\mathsf{A} : X \rightarrow Y}$.

Linear operator ${\mathsf{A}}$ between Hilbert spaces ${X}$ and ${Y}$, and its adjoint ${\mathsf{A}^\mathrm{T}}$.

The adjoint of ${\mathsf{A}}$ is defined as the operator ${\mathsf{A}^{*} : Y\rightarrow X}$ such that

$\displaystyle \langle \mathsf{A} f , g \rangle = \langle f , \mathsf{A}^{*}g \rangle$

For real-valued functions, we write ${ \mathsf{A}^{*} = \mathsf{A}^\mathrm{T}}$. In the finite-dimensional case ${\langle f, g \rangle = \sum_{i} f_i g_i}$ and ${\mathsf{A}}$ may be represented by a matrix ${\mathsf{A}_{ij}}$. Its adjoint is the transformed matrix ${\mathsf{A}^\mathrm{T}_{ij} = \mathsf{A}_{ji}}$.

The differential operator ${\mathrm{d}/\mathrm{d}x}$ for functions on a bounded interval can be represented by a skew-symmetric matrix

$\displaystyle \mathsf{D} = \frac{1}{2\Delta x} \left[ \begin{matrix} 0 & 1 & 0 & 0 & \dots \\ -1 & 0 & 1 & 0 & \dots \\ 0 & -1 & 0 & 1 & \dots \\ 0 & 0 & -1 & 0 & \dots \\ \vdots & \vdots & \vdots & \vdots & \ddots \end{matrix} \right]$

with transform ${\mathsf{D}^\mathrm{T} = - \mathsf{D}}$.

In the continuous case, we have no matrix representation and we use integration by parts to obtain the adjoint:

$\displaystyle \langle \mathsf{D} f, g \rangle = \int_a^b \mathsf{D}(f) g\,\mathrm{d}x = \int_a^b [ \mathsf{D}(f g) - f \mathsf{D}g] \,\mathrm{d}x = [ f g]_a^b - \int_a^b f \mathsf{D}g \,\mathrm{d}x$

Assuming the boundary term vanishes, this means that

$\displaystyle \langle \mathsf{D} f, g \rangle = - \langle f , \mathsf{D} g \rangle$

so that ${\mathsf{D}^\mathrm{T} = - \mathsf{D}}$, just as in the finite-dimensional case.

Operators on Scalar and Vector Fields

We let ${{\cal S} = S(\mathbb{R}^3,\mathbb{R})}$ be the space of (smooth) scalar fields over ${\mathbb{R}^3}$ and ${{\cal V} = \boldsymbol{V}(\mathbb{R}^3,\mathbb{R}^3)}$ the space of smooth vector-valued fields on ${\mathbb{R}^3}$. The vector operators ${\mathbf{grad}}$, ${\mathrm{div}}$ and ${\mathbf{curl}}$ define mappings between these function spaces, as shown in the diagram:

Vector operators grad, div and curl, mapping between the function spaces ${{S} }$ and ${\boldsymbol{V}}$.

We note that

$\displaystyle \mathbf{grad} : {\cal S}\rightarrow{\cal V}\,, \qquad \mathrm{div} : {\cal V}\rightarrow{\cal S}\,, \qquad \mathbf{curl} : {\cal V}\rightarrow{\cal V} \,.$

Clearly, some compositions of these operators are well-defined while others are not. The legal combinations of two operators (right operator first) are

$\displaystyle \begin{array}{rcl} \nabla^2 = \mathrm{div}\ \mathbf{grad} &:& {\cal S}\rightarrow{\cal S} \\ \mathbf{grad}\ \mathrm{div} &:& {\cal V}\rightarrow{\cal V} \\ \mathbf{0} \equiv \mathbf{c url}\ \mathbf{grad} &:& {\cal S}\rightarrow{\cal V} \\ \mathrm{0} \equiv \mathrm{div}\ \mathbf{curl} &:& {\cal V}\rightarrow{ \cal S} \\ \mathbf{curl}\ \mathbf{curl} &:& {\cal V}\rightarrow{\cal V} \end{array}$

The vector Laplacian, mapping ${\cal V}$ to itself, is defined by ${\boldsymbol{\nabla}^2\mathbf{A} = \mathbf{grad\ }\mathrm{div}\mathrm{A} - \mathbf{curl\ curl\ A}}$.

For the function spaces ${\cal S}$ and ${\cal V}$ we define the inner products by integration:

$\displaystyle \langle \phi , \chi \rangle = \int\!\!\!\!\int\!\!\!\!\int_{V} \phi\,\chi\,\mathrm{d} v \qquad\mbox{and}\qquad \langle \mathbf{A} , \mathbf{B} \rangle = \int\!\!\!\!\int\!\!\!\!\int_{V} \mathbf{A \cdot B} \,\mathrm{d}v$

Let us start with the divergence operator, and form ${\langle \phi, \mathrm{div}\ \mathbf{A} \rangle}$:

$\displaystyle \int\!\!\!\!\int\!\!\!\!\int_V \phi\,\boldsymbol{\nabla\cdot}\mathbf{A}\,\mathrm{d}v = \int\!\!\!\!\int\!\!\!\!\int_V [\boldsymbol{\nabla\cdot}(\phi \mathbf{A}) - \boldsymbol{\nabla}\phi\mathbf{\cdot A}]\,\mathrm{d}v$  $\displaystyle = \int\!\!\!\!\int_S \phi\mathbf{A\cdot n}\,\mathrm{d}\sigma - \int\!\!\!\!\int\!\!\!\!\int_V \boldsymbol{\nabla}\phi\mathbf{\cdot A}\,\mathrm{d}v$

Assuming that the boundary term (the surface integral) vanishes, this gives

$\displaystyle \int\!\!\! \!\int\!\!\!\!\int_V \phi\,\boldsymbol{\nabla\cdot }\mathbf{A}\,\mathrm{d}v = - \int\!\!\!\!\int\!\!\!\!\int_V \boldsymbol{\nabla}\phi\mathbf{\cdot A}\,\mathrm{d}v$

which may be written

$\displaystyle \langle \phi, (\mathrm{div}) \mathbf{A} \rangle = \langle (\mathbf{-grad})\phi, \mathbf{A} \rangle \,.$

This gives us the transpose ${(\mathrm{div})^\mathrm{T} = - \mathbf{grad}}$, and also ${(\mathbf{grad})^\mathrm{T} = -\mathrm{div}}$ (these are reminiscent of ${\mathsf{D}^\mathrm{T} = - \mathsf{D}}$ above). We also note that

$\displaystyle (\nabla^2)^\mathrm{T} = (\mathrm{div}\ \mathbf{grad})^\mathrm{T} = \mathbf{grad}^\mathrm{T}\ \mathrm{div}^\mathrm{T} \\ = (-\mathrm{div}) (- \mathbf{grad}) = \mathrm{div}\ \mathbf{grad} = \nabla^2$

so the Laplacian operator is self-adjoint.

Now let us look at the curl operator, and form ${\langle \mathbf{A}, \mathbf{curl}\ \mathbf{A} \rangle}$. We use the vector identity

$\displaystyle \boldsymbol{\nabla\cdot}(\mathbf{A\times B}) = (\boldsymbol{\nabla\times}\mathbf{A})\mathbf{\cdot B} - \mathbf{A\cdot} (\boldsymbol{\nabla\times}\mathbf{B})$

Assuming that the divergence term integrates to zero, this gives

$\displaystyle \int\!\!\!\!\int\!\!\!\!\int_V \mathbf{A}\boldsymbol{\cdot\nabla\times}\mathbf{B}\,\mathrm{d}v = \int\!\!\!\!\int\!\!\!\!\int_V \boldsymbol{\nabla \times}\mathbf{A}\ \mathbf{\cdot\ B}\,\mathrm{d}v$

which may be written

$\displaystyle \langle \mathrm{A}, \mathbf{curl}\ \mathbf{B} \rangle = \langle \mathbf{curl}\ \mathbf{A}, \mathbf{B} \rangle \,.$

This gives us the transpose ${(\mathbf{curl})^\mathrm{T} = \mathbf{curl}}$, showing that the ${\mathbf {curl}}$ operator is symmetric.

Finally, using ${(\mathrm{div})^\mathrm{T} = - \mathbf{grad}}$ and ${(\mathbf{curl})^\mathrm{T} = \mathbf{curl}}$, we have

$\displaystyle (\mathbf{curl}\ \mathbf{grad})^\mathrm{T} = \mathbf{grad}^\mathrm{T} \mathbf{curl}^\mathrm{T} = - \mathrm{div}\ \mathbf{curl}$

so the two vector identities with which we started,

$\displaystyle \begin{array}{rcl} \boldsymbol{\nabla\times\nabla}\chi &= \mathbf{curl}\ \mathbf{grad}\ \chi &\equiv 0\,, \quad \mbox{for all scalar functions\ }\chi \\ \boldsymbol{\nabla\cdot\nabla\times\psi} &= \mathrm{div}\ \mathbf{curl}\ \boldsymbol{\psi} &\equiv 0\,, \quad \mbox{for all vector functions\ }\boldsymbol{\psi} \,, \end{array}$

are actually adjoints of each other.

$\displaystyle \langle \mathbf{curl}\ \mathbf{grad}\ \phi, \mathbf{A} \rangle = \langle \phi, \mathrm{div}\ \mathbf{curl}\ \mathbf{A} \rangle \,.$
${\bullet}$ Strang, Gilbert, 1986: Introduction to Applied Mathematics. Wellesley-Cambridge Press, 758pp. ISBN: 0-961-40880-4.