Proofs Chapter 3: Divergence Formulas

Linearity, product rules, and combinations with gradient (21)-(28)

3. Divergence Formulas

This chapter proves identities (21)-(28) involving the divergence $\nabla \cdot \mathbf{A} = \displaystyle\sum_{i} \dfrac{\partial A_i}{\partial x_i}$ of a vector field. Each proof proceeds by component expansion combined with linearity of partial differentiation and the Leibniz product rule, drawing on the gradient identities established in Chapter 2 and the cyclic property of the scalar triple product (5) where needed.

Roadmap

This chapter is organized as follows.

  • 3.1 Linearity (21-22): distributivity over sums and scalar multiples
  • 3.2 Product Rules (23-24): Leibniz-type rules for scalar-vector products and cross products
  • 3.3 Combinations with Gradient (25-28): cross products of gradients and the differential forms of Green's first and second identities
Geometric meaning of the divergence: the net flux through the faces of a small box equals the divergence dV inflow A·n dS outflow A·n dS x z y net outward flux ÷ dV → ∇·A (source strength)
Figure 1. Geometric meaning of the divergence. For the flux $\mathbf{A}\cdot\mathbf{n}\,dS$ entering and leaving each face of an infinitesimal box $dV$, the amount by which the outflow (blue) exceeds the inflow (orange) is the net source; dividing it by the volume and taking the limit gives the divergence $\nabla\cdot\mathbf{A}$. Conversely, if the inflow exceeds the outflow then $\nabla\cdot\mathbf{A}<0$ (a sink), and if they balance then $\nabla\cdot\mathbf{A}=0$ (incompressible). Formulas (21)-(28) of this chapter are identities for this quantity.

3.1 Linearity

The divergence is a linear combination of partial derivatives, so it distributes over sums and scalar multiples.

(21) Linearity of the divergence (sum)

Formula: $\nabla \cdot (\mathbf{A} + \mathbf{B}) = \nabla \cdot \mathbf{A} + \nabla \cdot \mathbf{B}$
Conditions: $\mathbf{A}, \mathbf{B}$ are $C^1$ vector fields.
Proof

Expand by the definition of the divergence in Cartesian components.

\begin{equation}\nabla \cdot (\mathbf{A} + \mathbf{B}) = \displaystyle\sum_{i=1}^{3} \dfrac{\partial (A_i + B_i)}{\partial x_i} \label{eq:3-21-1}\end{equation}

By the linearity of partial differentiation, the derivative of a sum equals the sum of derivatives.

\begin{equation}\displaystyle\sum_{i=1}^{3} \dfrac{\partial (A_i + B_i)}{\partial x_i} = \displaystyle\sum_{i=1}^{3} \dfrac{\partial A_i}{\partial x_i} + \displaystyle\sum_{i=1}^{3} \dfrac{\partial B_i}{\partial x_i} \label{eq:3-21-2}\end{equation}

The two sums on the right are exactly $\nabla \cdot \mathbf{A}$ and $\nabla \cdot \mathbf{B}$.

\begin{equation}\nabla \cdot (\mathbf{A} + \mathbf{B}) = \nabla \cdot \mathbf{A} + \nabla \cdot \mathbf{B} \label{eq:3-21-3}\end{equation}

Remark: Linearity of partial differentiation is just the linearity of the single-variable derivative (1.24) applied componentwise.

(22) Scalar multiple

Formula: $\nabla \cdot (c\,\mathbf{A}) = c\,\nabla \cdot \mathbf{A}$
Conditions: $c$ is a constant (independent of position); $\mathbf{A}$ is a $C^1$ vector field.
Proof

Substitute into the definition of the divergence.

\begin{equation}\nabla \cdot (c\,\mathbf{A}) = \displaystyle\sum_{i=1}^{3} \dfrac{\partial (c\,A_i)}{\partial x_i} \label{eq:3-22-1}\end{equation}

Because $c$ is constant, it factors out of the partial derivative.

\begin{equation}\displaystyle\sum_{i=1}^{3} \dfrac{\partial (c\,A_i)}{\partial x_i} = \displaystyle\sum_{i=1}^{3} c\,\dfrac{\partial A_i}{\partial x_i} = c \displaystyle\sum_{i=1}^{3} \dfrac{\partial A_i}{\partial x_i} = c\,\nabla \cdot \mathbf{A} \label{eq:3-22-2}\end{equation}

Remark: When $c$ depends on $\mathbf{x}$ one must use formula (23) instead; (22) does not apply.

3.2 Product Rules

Divergences of scalar-vector products and of cross products of two vector fields satisfy Leibniz-type rules.

(23) Divergence of a scalar-vector product

Formula: $\nabla \cdot (f\mathbf{A}) = f(\nabla \cdot \mathbf{A}) + \mathbf{A} \cdot (\nabla f)$
Conditions: $f$ is a $C^1$ scalar field, $\mathbf{A}$ is a $C^1$ vector field.
Proof

Expand by the definition of the divergence in components.

\begin{equation}\nabla \cdot (f\mathbf{A}) = \displaystyle\sum_{i=1}^{3} \dfrac{\partial (f\,A_i)}{\partial x_i} \label{eq:3-23-1}\end{equation}

Apply the single-variable product rule (1.25) componentwise.

\begin{equation}\dfrac{\partial (f\,A_i)}{\partial x_i} = f\,\dfrac{\partial A_i}{\partial x_i} + A_i\,\dfrac{\partial f}{\partial x_i} \label{eq:3-23-2}\end{equation}

Sum over $i$.

\begin{equation}\nabla \cdot (f\mathbf{A}) = \displaystyle\sum_{i=1}^{3} f\,\dfrac{\partial A_i}{\partial x_i} + \displaystyle\sum_{i=1}^{3} A_i\,\dfrac{\partial f}{\partial x_i} \label{eq:3-23-3}\end{equation}

In the first sum, $f$ is independent of $i$ and factors out, yielding $f\,\nabla \cdot \mathbf{A}$. The second sum is the dot product of $\mathbf{A}$ and $\nabla f$.

\begin{equation}\nabla \cdot (f\mathbf{A}) = f\,(\nabla \cdot \mathbf{A}) + \mathbf{A} \cdot (\nabla f) \label{eq:3-23-4}\end{equation}

Remark: This is the divergence Leibniz rule and underlies Green's first identity (formula 26) as well as the Helmholtz decomposition.

(24) Divergence of a cross product

Formula: $\nabla \cdot (\mathbf{A} \times \mathbf{B}) = \mathbf{B} \cdot (\nabla \times \mathbf{A}) - \mathbf{A} \cdot (\nabla \times \mathbf{B})$
Conditions: $\mathbf{A}, \mathbf{B}$ are $C^1$ vector fields.
Proof

Write the cross product in components using the Levi-Civita symbol $\varepsilon_{ijk}$ (defined in Chapter 1).

\begin{equation}(\mathbf{A} \times \mathbf{B})_i = \displaystyle\sum_{j,k} \varepsilon_{ijk}\,A_j\,B_k \label{eq:3-24-1}\end{equation}

Substitute into the definition of the divergence.

\begin{equation}\nabla \cdot (\mathbf{A} \times \mathbf{B}) = \displaystyle\sum_{i,j,k} \varepsilon_{ijk}\,\dfrac{\partial (A_j\,B_k)}{\partial x_i} \label{eq:3-24-2}\end{equation}

Apply the product rule.

\begin{equation}\nabla \cdot (\mathbf{A} \times \mathbf{B}) = \displaystyle\sum_{i,j,k} \varepsilon_{ijk}\!\left( B_k\,\dfrac{\partial A_j}{\partial x_i} + A_j\,\dfrac{\partial B_k}{\partial x_i} \right) \label{eq:3-24-3}\end{equation}

First term. Reorder the summations and factor $B_k$ outside.

\begin{equation}\displaystyle\sum_{i,j,k} \varepsilon_{ijk}\,B_k\,\dfrac{\partial A_j}{\partial x_i} = \displaystyle\sum_{k} B_k \displaystyle\sum_{i,j} \varepsilon_{ijk}\,\dfrac{\partial A_j}{\partial x_i} \label{eq:3-24-4}\end{equation}

Since $\varepsilon_{ijk} = \varepsilon_{kij}$ (cyclic permutation preserves the sign), the inner sum $\displaystyle\sum_{i,j} \varepsilon_{kij}\,\partial A_j / \partial x_i$ is the $k$-th component of the curl $\nabla \times \mathbf{A}$.

\begin{equation}\displaystyle\sum_{i,j,k} \varepsilon_{ijk}\,B_k\,\dfrac{\partial A_j}{\partial x_i} = \displaystyle\sum_{k} B_k\,(\nabla \times \mathbf{A})_k = \mathbf{B} \cdot (\nabla \times \mathbf{A}) \label{eq:3-24-5}\end{equation}

Second term. Factor $A_j$ outside in the same way.

\begin{equation}\displaystyle\sum_{i,j,k} \varepsilon_{ijk}\,A_j\,\dfrac{\partial B_k}{\partial x_i} = \displaystyle\sum_{j} A_j \displaystyle\sum_{i,k} \varepsilon_{ijk}\,\dfrac{\partial B_k}{\partial x_i} \label{eq:3-24-6}\end{equation}

Use the antisymmetry $\varepsilon_{ijk} = -\varepsilon_{ikj}$. Swapping the dummy indices $i \leftrightarrow k$ gives

\begin{equation}\displaystyle\sum_{i,k} \varepsilon_{ijk}\,\dfrac{\partial B_k}{\partial x_i} = -\displaystyle\sum_{k,i} \varepsilon_{jki}\,\dfrac{\partial B_k}{\partial x_i} \cdot (\text{cyclic rewrite}) = -(\nabla \times \mathbf{B})_j \label{eq:3-24-7}\end{equation}

Hence

\begin{equation}\displaystyle\sum_{i,j,k} \varepsilon_{ijk}\,A_j\,\dfrac{\partial B_k}{\partial x_i} = -\displaystyle\sum_{j} A_j\,(\nabla \times \mathbf{B})_j = -\,\mathbf{A} \cdot (\nabla \times \mathbf{B}) \label{eq:3-24-8}\end{equation}

Substituting $\eqref{eq:3-24-5}$ and $\eqref{eq:3-24-8}$ into $\eqref{eq:3-24-3}$ yields

\begin{equation}\nabla \cdot (\mathbf{A} \times \mathbf{B}) = \mathbf{B} \cdot (\nabla \times \mathbf{A}) - \mathbf{A} \cdot (\nabla \times \mathbf{B}) \label{eq:3-24-9}\end{equation}

Remark: The sign difference between the two terms reflects the antisymmetry $\varepsilon_{ijk} = -\varepsilon_{ikj}$. This identity is used, for instance, to compute the divergence of the Poynting vector $\mathbf{S} = \mathbf{E} \times \mathbf{H}$.

3.3 Combinations with Gradient

Identities mixing gradients with the divergence introduce the Laplacian and lead to Green's identities, the cornerstone tools of partial differential equation theory.

(25) Divergence of a cross product of gradients

Formula: $\nabla \cdot (\nabla f \times \nabla g) = 0$
Conditions: $f, g$ are $C^2$ scalar fields.
Proof

Apply formula (24) with $\mathbf{A} = \nabla f$ and $\mathbf{B} = \nabla g$.

\begin{equation}\nabla \cdot (\nabla f \times \nabla g) = \nabla g \cdot (\nabla \times \nabla f) - \nabla f \cdot (\nabla \times \nabla g) \label{eq:3-25-1}\end{equation}

The curl of a gradient vanishes (formula (38)): $\nabla \times \nabla f = \mathbf{0}$ and $\nabla \times \nabla g = \mathbf{0}$ for any $C^2$ scalar fields.

\begin{equation}\nabla \cdot (\nabla f \times \nabla g) = \nabla g \cdot \mathbf{0} - \nabla f \cdot \mathbf{0} = 0 \label{eq:3-25-2}\end{equation}

Remark: The identity says that $\nabla f \times \nabla g$ is solenoidal (divergence-free). Geometrically, the field tangent to the intersection curves of the level surfaces $f = \text{const}$ and $g = \text{const}$ has no sources or sinks.

(26) Differential form of Green's first identity

Formula: $\nabla \cdot (f\,\nabla g) = f\,\nabla^2 g + \nabla f \cdot \nabla g$
Conditions: $f$ is $C^1$, $g$ is $C^2$.
Proof

Apply formula (23) with $\mathbf{A} = \nabla g$.

\begin{equation}\nabla \cdot (f\,\nabla g) = f\,(\nabla \cdot \nabla g) + (\nabla g) \cdot (\nabla f) \label{eq:3-26-1}\end{equation}

Since $\nabla \cdot \nabla g = \nabla^2 g$ (the Laplacian, formula (36)) and the dot product is symmetric, $(\nabla g) \cdot (\nabla f) = \nabla f \cdot \nabla g$.

\begin{equation}\nabla \cdot (f\,\nabla g) = f\,\nabla^2 g + \nabla f \cdot \nabla g \label{eq:3-26-2}\end{equation}

Remark: Integrating this identity over a domain $\Omega$ and applying the divergence theorem yields Green's first identity $\displaystyle\int_{\Omega}\!\bigl( f\,\nabla^2 g + \nabla f \cdot \nabla g \bigr)\,dV = \displaystyle\oint_{\partial \Omega} f\,(\nabla g \cdot \mathbf{n})\,dS$.

(27) Differential form of Green's second identity

Formula: $\nabla \cdot (f\,\nabla g - g\,\nabla f) = f\,\nabla^2 g - g\,\nabla^2 f$
Conditions: $f, g$ are $C^2$ scalar fields.
Proof

Split the left-hand side using linearity of the divergence (21).

\begin{equation}\nabla \cdot (f\,\nabla g - g\,\nabla f) = \nabla \cdot (f\,\nabla g) - \nabla \cdot (g\,\nabla f) \label{eq:3-27-1}\end{equation}

Apply formula (26) to each term.

\begin{equation}\nabla \cdot (f\,\nabla g) = f\,\nabla^2 g + \nabla f \cdot \nabla g \label{eq:3-27-2}\end{equation}

\begin{equation}\nabla \cdot (g\,\nabla f) = g\,\nabla^2 f + \nabla g \cdot \nabla f \label{eq:3-27-3}\end{equation}

Subtract $\eqref{eq:3-27-3}$ from $\eqref{eq:3-27-2}$. By symmetry of the dot product, $\nabla f \cdot \nabla g = \nabla g \cdot \nabla f$, so the cross terms cancel.

\begin{equation}\nabla \cdot (f\,\nabla g - g\,\nabla f) = f\,\nabla^2 g - g\,\nabla^2 f + \underbrace{\nabla f \cdot \nabla g - \nabla g \cdot \nabla f}_{= 0} \label{eq:3-27-4}\end{equation}

\begin{equation}\nabla \cdot (f\,\nabla g - g\,\nabla f) = f\,\nabla^2 g - g\,\nabla^2 f \label{eq:3-27-5}\end{equation}

Remark: Integrating over a domain $\Omega$ produces Green's second identity $\displaystyle\int_{\Omega}\!\bigl( f\,\nabla^2 g - g\,\nabla^2 f \bigr)\,dV = \displaystyle\oint_{\partial \Omega}\!\bigl( f\,\nabla g - g\,\nabla f \bigr) \cdot \mathbf{n}\,dS$, on which the Green's-function method for the Poisson equation is built.

(28) Divergence of a scalar triple product

Formula: $\nabla \cdot (f\,\nabla g \times \nabla h) = \nabla f \cdot (\nabla g \times \nabla h)$
Conditions: $f$ is $C^1$, $g, h$ are $C^2$ scalar fields.
Proof

Set $\mathbf{A} = \nabla g \times \nabla h$ and apply formula (23).

\begin{equation}\nabla \cdot (f\,\mathbf{A}) = f\,(\nabla \cdot \mathbf{A}) + \mathbf{A} \cdot (\nabla f) \label{eq:3-28-1}\end{equation}

First term. By formula (25), $\nabla \cdot (\nabla g \times \nabla h) = 0$.

\begin{equation}f\,(\nabla \cdot \mathbf{A}) = f\,(\nabla \cdot (\nabla g \times \nabla h)) = 0 \label{eq:3-28-2}\end{equation}

Second term. Use the cyclic property of the scalar triple product (formula (5)): $\mathbf{A} \cdot (\mathbf{B} \times \mathbf{C}) = \mathbf{B} \cdot (\mathbf{C} \times \mathbf{A}) = \mathbf{C} \cdot (\mathbf{A} \times \mathbf{B})$.

\begin{equation}\mathbf{A} \cdot (\nabla f) = (\nabla g \times \nabla h) \cdot (\nabla f) = \nabla f \cdot (\nabla g \times \nabla h) \label{eq:3-28-3}\end{equation}

Substituting $\eqref{eq:3-28-2}$ and $\eqref{eq:3-28-3}$ into $\eqref{eq:3-28-1}$,

\begin{equation}\nabla \cdot (f\,\nabla g \times \nabla h) = 0 + \nabla f \cdot (\nabla g \times \nabla h) = \nabla f \cdot (\nabla g \times \nabla h) \label{eq:3-28-4}\end{equation}

Remark: The right-hand side is the Jacobian $\dfrac{\partial(f, g, h)}{\partial(x, y, z)}$. The condition $\nabla f \cdot (\nabla g \times \nabla h) \neq 0$ characterizes functional independence of the three scalar fields.

References

  • Schey, H. M. (2005). Div, Grad, Curl, and All That: An Informal Text on Vector Calculus (4th ed.). W. W. Norton.
  • Marsden, J. E., & Tromba, A. J. (2011). Vector Calculus (6th ed.). W. H. Freeman.
  • Vector calculus identities - Wikipedia

Frequently Asked Questions

What are the product rules for divergence?

Key formulas: $\nabla\cdot(f\mathbf{F})=f(\nabla\cdot\mathbf{F})+(\nabla f)\cdot\mathbf{F}$ for scalar-vector products, and $\nabla\cdot(\mathbf{F}\times\mathbf{G})=\mathbf{G}\cdot(\nabla\times\mathbf{F})-\mathbf{F}\cdot(\nabla\times\mathbf{G})$ for cross products. These enable integration by parts in multiple dimensions, for instance in deriving Green's identities.

What is a divergence-free (solenoidal) vector field?

A vector field $\mathbf{F}$ with $\nabla\cdot\mathbf{F}=0$ is called solenoidal or divergence-free. Incompressible fluids satisfy $\nabla\cdot\mathbf{v}=0$. By the divergence theorem, the net flux through any closed surface is zero for such fields. The curl of any vector field is always solenoidal: $\nabla\cdot(\nabla\times\mathbf{A})=0$.

What does the divergence theorem state and why is it important?

The divergence theorem states $\oiint_S \mathbf{F}\cdot d\mathbf{S}=\iiint_V (\nabla\cdot\mathbf{F})\,dV$: the total flux through a closed surface equals the volume integral of divergence inside. This underlies Gauss's law in electromagnetism ($\oiint_S \mathbf{E}\cdot d\mathbf{S}=Q/\varepsilon_0$) and converts surface integrals to volume integrals in engineering calculations.