Proofs Chapter 7: The (A·∇) Operator

Discussion

The symbol $(\mathbf{A} \cdot \nabla)$ should be read as the inner product $\mathbf{A} \cdot \nabla$ regarded as a single operator. Formally substituting the components of $\nabla = (\partial/\partial x, \partial/\partial y, \partial/\partial z)$ and $\mathbf{A} = (A_x, A_y, A_z)$ gives

\begin{equation}\mathbf{A} \cdot \nabla = A_x \dfrac{\partial}{\partial x} + A_y \dfrac{\partial}{\partial y} + A_z \dfrac{\partial}{\partial z} \label{eq:7-1-1}\end{equation}

This is a scalar-valued first-order linear differential operator that can act on either a scalar field or a vector field.

Well-definedness. The right-hand side of $\eqref{eq:7-1-1}$ is a finite sum of linear differential operators, so its value is uniquely determined whenever the target $\mathbf{B}$ is differentiable. We define $(\mathbf{A} \cdot \nabla)\mathbf{B}$ as the vector field obtained by applying the scalar operator $\mathbf{A} \cdot \nabla$ component-by-component to $\mathbf{B}$.

Relation to the directional derivative. Writing $\hat{\mathbf{n}} = \mathbf{A}/|\mathbf{A}|$ for the unit vector along $\mathbf{A}$, we have

\begin{equation}(\mathbf{A} \cdot \nabla)\mathbf{B} = |\mathbf{A}| \, (\hat{\mathbf{n}} \cdot \nabla)\mathbf{B} \label{eq:7-1-2}\end{equation}

The quantity $(\hat{\mathbf{n}} \cdot \nabla)\mathbf{B}$ is the directional derivative of $\mathbf{B}$ along $\hat{\mathbf{n}}$, so $(\mathbf{A} \cdot \nabla)\mathbf{B}$ equals that directional derivative scaled by the magnitude $|\mathbf{A}|$.

Proof

Prerequisites: definition (58); the components of a vector field are differentiated independently.

From definition (58),

\begin{equation}(\mathbf{A} \cdot \nabla)\mathbf{B} = \left(A_x \dfrac{\partial}{\partial x} + A_y \dfrac{\partial}{\partial y} + A_z \dfrac{\partial}{\partial z}\right)\mathbf{B} \label{eq:7-2-1}\end{equation}

The right-hand side acts on the vector field $\mathbf{B} = (B_x, B_y, B_z)$, and its $i$-th component is therefore

\begin{equation}\bigl[(\mathbf{A} \cdot \nabla)\mathbf{B}\bigr]_i = \left(A_x \dfrac{\partial}{\partial x} + A_y \dfrac{\partial}{\partial y} + A_z \dfrac{\partial}{\partial z}\right) B_i \label{eq:7-2-2}\end{equation}

Linear differential operators distribute over their argument, so $\eqref{eq:7-2-2}$ expands to

\begin{equation}\bigl[(\mathbf{A} \cdot \nabla)\mathbf{B}\bigr]_i = A_x \dfrac{\partial B_i}{\partial x} + A_y \dfrac{\partial B_i}{\partial y} + A_z \dfrac{\partial B_i}{\partial z} \label{eq:7-2-3}\end{equation}

which is the desired identity. With the Einstein summation convention this can be written compactly as $\sum_j A_j \,\partial B_i/\partial x_j$.

Proof

Prerequisites: definition (58); the gradient $\nabla f = (\partial f/\partial x, \partial f/\partial y, \partial f/\partial z)$; the definition of the inner product.

Expand the left-hand side using definition (58).

\begin{equation}(\mathbf{A} \cdot \nabla)f = \left(A_x \dfrac{\partial}{\partial x} + A_y \dfrac{\partial}{\partial y} + A_z \dfrac{\partial}{\partial z}\right) f = A_x \dfrac{\partial f}{\partial x} + A_y \dfrac{\partial f}{\partial y} + A_z \dfrac{\partial f}{\partial z} \label{eq:7-3-1}\end{equation}

The right-hand expression is exactly the dot product of $\mathbf{A} = (A_x, A_y, A_z)$ with $\nabla f = (\partial f/\partial x, \partial f/\partial y, \partial f/\partial z)$:

\begin{equation}A_x \dfrac{\partial f}{\partial x} + A_y \dfrac{\partial f}{\partial y} + A_z \dfrac{\partial f}{\partial z} = \mathbf{A} \cdot \nabla f \label{eq:7-3-2}\end{equation}

From $\eqref{eq:7-3-1}$ and $\eqref{eq:7-3-2}$ the identity $(\mathbf{A} \cdot \nabla)f = \mathbf{A} \cdot \nabla f$ follows.

Proof

Prerequisites: the Levi-Civita symbol $\varepsilon_{ijk}$; the $\varepsilon$-$\delta$ identity $\displaystyle\sum_k \varepsilon_{ijk}\varepsilon_{klm} = \delta_{il}\delta_{jm} - \delta_{im}\delta_{jl}$; the component formulas $[\mathbf{a} \times \mathbf{b}]_i = \varepsilon_{ijk} a_j b_k$ and $[\nabla \times \mathbf{B}]_k = \varepsilon_{klm} \,\partial B_m/\partial x_l$. Einstein summation is in force.

(i) Component expansion of the LHS. Expand the cross product in two stages.

\begin{equation}[\mathbf{A} \times (\nabla \times \mathbf{B})]_i = \varepsilon_{ijk} A_j \, [\nabla \times \mathbf{B}]_k = \varepsilon_{ijk} \varepsilon_{klm} A_j \dfrac{\partial B_m}{\partial x_l} \label{eq:7-4-1}\end{equation}

(ii) Apply the $\varepsilon$-$\delta$ identity. Substituting $\varepsilon_{ijk} \varepsilon_{klm} = \delta_{il}\delta_{jm} - \delta_{im}\delta_{jl}$ into $\eqref{eq:7-4-1}$ yields

\begin{equation}[\mathbf{A} \times (\nabla \times \mathbf{B})]_i = (\delta_{il}\delta_{jm} - \delta_{im}\delta_{jl}) A_j \dfrac{\partial B_m}{\partial x_l} \label{eq:7-4-2}\end{equation}

(iii) Contract the Kronecker deltas. $\delta_{il}$ forces $l = i$ and $\delta_{jm}$ forces $m = j$, so the first piece becomes

\begin{equation}\delta_{il}\delta_{jm} A_j \dfrac{\partial B_m}{\partial x_l} = A_j \dfrac{\partial B_j}{\partial x_i} \label{eq:7-4-3}\end{equation}

Similarly $\delta_{im}\delta_{jl}$ enforces $m = i$ and $l = j$, giving

\begin{equation}\delta_{im}\delta_{jl} A_j \dfrac{\partial B_m}{\partial x_l} = A_j \dfrac{\partial B_i}{\partial x_j} \label{eq:7-4-4}\end{equation}

(iv) Combine. From $\eqref{eq:7-4-2}$, $\eqref{eq:7-4-3}$, $\eqref{eq:7-4-4}$:

\begin{equation}[\mathbf{A} \times (\nabla \times \mathbf{B})]_i = A_j \dfrac{\partial B_j}{\partial x_i} - A_j \dfrac{\partial B_i}{\partial x_j} \label{eq:7-4-5}\end{equation}

(v) Identify each piece. The first sum $\displaystyle\sum_j A_j \,\partial B_j/\partial x_i$ is exactly $[(\nabla \mathbf{B}) \cdot \mathbf{A}]_i$ by the convention stated in the conditions. The second sum $\displaystyle\sum_j A_j \,\partial B_i/\partial x_j$ equals $[(\mathbf{A} \cdot \nabla)\mathbf{B}]_i$ by (59). Hence

\begin{equation}[\mathbf{A} \times (\nabla \times \mathbf{B})]_i = [(\nabla \mathbf{B}) \cdot \mathbf{A}]_i - [(\mathbf{A} \cdot \nabla)\mathbf{B}]_i \label{eq:7-4-6}\end{equation}

Since $i$ was arbitrary, we obtain the vector identity $\mathbf{A} \times (\nabla \times \mathbf{B}) = (\nabla \mathbf{B}) \cdot \mathbf{A} - (\mathbf{A} \cdot \nabla)\mathbf{B}$, as required.

Proof

Prerequisites: equation $\eqref{eq:7-4-5}$ from the proof of (61).

Steps (i)–(iv) of the proof of (61) produced

\begin{equation}[\mathbf{A} \times (\nabla \times \mathbf{B})]_i = \displaystyle\sum_j A_j \dfrac{\partial B_j}{\partial x_i} - \displaystyle\sum_j A_j \dfrac{\partial B_i}{\partial x_j} \label{eq:7-5-1}\end{equation}

which is precisely identity (62). In short, (62) is the component form of the vector identity (61).

Proof

Prerequisites: the gradient-of-dot-product formula (16): $\nabla(\mathbf{A} \cdot \mathbf{B}) = (\mathbf{B} \cdot \nabla)\mathbf{A} + (\mathbf{A} \cdot \nabla)\mathbf{B} + \mathbf{B} \times (\nabla \times \mathbf{A}) + \mathbf{A} \times (\nabla \times \mathbf{B})$, which we take as already established.

Start from (16):

\begin{equation}\nabla(\mathbf{A} \cdot \mathbf{B}) = (\mathbf{B} \cdot \nabla)\mathbf{A} + (\mathbf{A} \cdot \nabla)\mathbf{B} + \mathbf{B} \times (\nabla \times \mathbf{A}) + \mathbf{A} \times (\nabla \times \mathbf{B}) \label{eq:7-6-1}\end{equation}

Subtracting $\mathbf{A} \times (\nabla \times \mathbf{B}) + \mathbf{B} \times (\nabla \times \mathbf{A})$ from both sides gives

\begin{equation}\nabla(\mathbf{A} \cdot \mathbf{B}) - \mathbf{A} \times (\nabla \times \mathbf{B}) - \mathbf{B} \times (\nabla \times \mathbf{A}) = (\mathbf{A} \cdot \nabla)\mathbf{B} + (\mathbf{B} \cdot \nabla)\mathbf{A} \label{eq:7-6-2}\end{equation}

which is exactly (63).

Proof

Prerequisites: a fluid-particle trajectory $\mathbf{x}(t) = (x(t), y(t), z(t))$ satisfies $d\mathbf{x}/dt = \mathbf{v}(\mathbf{x}(t), t)$; the multivariate chain rule; identity (60).

(i) Definition of the material derivative. The material derivative $Df/Dt$ is defined as the total derivative of the composite function $f(\mathbf{x}(t), t)$ along a fluid trajectory $\mathbf{x}(t)$:

\begin{equation}\dfrac{Df}{Dt} = \dfrac{d}{dt} f(\mathbf{x}(t), t) \label{eq:7-7-1}\end{equation}

(ii) Multivariate chain rule. Since $f$ is a function of the four variables $x, y, z, t$ and $x, y, z$ themselves depend on $t$, the chain rule gives

\begin{equation}\dfrac{d}{dt} f(\mathbf{x}(t), t) = \dfrac{\partial f}{\partial t} + \dfrac{\partial f}{\partial x} \dfrac{dx}{dt} + \dfrac{\partial f}{\partial y} \dfrac{dy}{dt} + \dfrac{\partial f}{\partial z} \dfrac{dz}{dt} \label{eq:7-7-2}\end{equation}

(iii) Substitute the velocity. By definition of the trajectory, $dx/dt = v_x$, $dy/dt = v_y$, $dz/dt = v_z$, so

\begin{equation}\dfrac{d}{dt} f(\mathbf{x}(t), t) = \dfrac{\partial f}{\partial t} + v_x \dfrac{\partial f}{\partial x} + v_y \dfrac{\partial f}{\partial y} + v_z \dfrac{\partial f}{\partial z} \label{eq:7-7-3}\end{equation}

(iv) Identify the advective term. The trailing terms of $\eqref{eq:7-7-3}$ form the dot product $\mathbf{v} \cdot \nabla f$, which by (60) equals $(\mathbf{v} \cdot \nabla)f$. Therefore

\begin{equation}\dfrac{Df}{Dt} = \dfrac{\partial f}{\partial t} + (\mathbf{v} \cdot \nabla)f \label{eq:7-7-4}\end{equation}

as claimed.

Proof

Prerequisites: the scalar material derivative (64); the components of a vector field have independent material derivatives; definition (58).

Each component $A_i$ of $\mathbf{A} = (A_x, A_y, A_z)$ is a scalar field on space-time, so identity (64) applies to it:

\begin{equation}\dfrac{DA_i}{Dt} = \dfrac{\partial A_i}{\partial t} + (\mathbf{v} \cdot \nabla) A_i \quad (i = x, y, z) \label{eq:7-8-1}\end{equation}

Stacking these three scalar equations into a single vector equation:

\begin{equation}\dfrac{D\mathbf{A}}{Dt} = \begin{pmatrix} DA_x/Dt \\ DA_y/Dt \\ DA_z/Dt \end{pmatrix} = \begin{pmatrix} \partial A_x/\partial t \\ \partial A_y/\partial t \\ \partial A_z/\partial t \end{pmatrix} + \begin{pmatrix} (\mathbf{v} \cdot \nabla) A_x \\ (\mathbf{v} \cdot \nabla) A_y \\ (\mathbf{v} \cdot \nabla) A_z \end{pmatrix} \label{eq:7-8-2}\end{equation}

The first vector on the right is $\partial \mathbf{A}/\partial t$, and the second one is $(\mathbf{v} \cdot \nabla)\mathbf{A}$ by the definition (58). Thus

\begin{equation}\dfrac{D\mathbf{A}}{Dt} = \dfrac{\partial \mathbf{A}}{\partial t} + (\mathbf{v} \cdot \nabla)\mathbf{A} \label{eq:7-8-3}\end{equation}

as required.

Proof

Prerequisites: the alternate form of the gradient of a dot product (63); $\mathbf{v} \cdot \mathbf{v} = |\mathbf{v}|^2$.

(i) Set $\mathbf{A} = \mathbf{B} = \mathbf{v}$ in (63):

\begin{equation}(\mathbf{v} \cdot \nabla)\mathbf{v} + (\mathbf{v} \cdot \nabla)\mathbf{v} = \nabla(\mathbf{v} \cdot \mathbf{v}) - \mathbf{v} \times (\nabla \times \mathbf{v}) - \mathbf{v} \times (\nabla \times \mathbf{v}) \label{eq:7-9-1}\end{equation}

(ii) Combine like terms:

\begin{equation}2 (\mathbf{v} \cdot \nabla)\mathbf{v} = \nabla(\mathbf{v} \cdot \mathbf{v}) - 2 \, \mathbf{v} \times (\nabla \times \mathbf{v}) \label{eq:7-9-2}\end{equation}

(iii) Divide both sides by 2:

\begin{equation}(\mathbf{v} \cdot \nabla)\mathbf{v} = \dfrac{1}{2} \nabla(\mathbf{v} \cdot \mathbf{v}) - \mathbf{v} \times (\nabla \times \mathbf{v}) \label{eq:7-9-3}\end{equation}

(iv) Substitute $\mathbf{v} \cdot \mathbf{v} = |\mathbf{v}|^2$. By linearity of the gradient $\dfrac{1}{2} \nabla(\mathbf{v} \cdot \mathbf{v}) = \nabla\!\left(\dfrac{|\mathbf{v}|^2}{2}\right)$, so

\begin{equation}(\mathbf{v} \cdot \nabla)\mathbf{v} = \nabla\!\left(\dfrac{|\mathbf{v}|^2}{2}\right) - \mathbf{v} \times (\nabla \times \mathbf{v}) \label{eq:7-9-4}\end{equation}

as desired.