Proofs Chapter 5: Second-Order Differential Identities

Proof

By the definition of gradient, the $i$-th component of $\nabla f$ is

\begin{equation}[\nabla f]_i = \dfrac{\partial f}{\partial x_i} \label{eq:5-36-1}\end{equation}

By the definition of divergence, for a vector field $\mathbf{V}$,

\begin{equation}\nabla \cdot \mathbf{V} = \displaystyle\sum_{i} \dfrac{\partial V_i}{\partial x_i} \label{eq:5-36-2}\end{equation}

Substituting the vector field $\nabla f$ from $\eqref{eq:5-36-1}$ into $\eqref{eq:5-36-2}$,

\begin{equation}\nabla \cdot (\nabla f) = \displaystyle\sum_{i} \dfrac{\partial}{\partial x_i}\!\left(\dfrac{\partial f}{\partial x_i}\right) = \displaystyle\sum_{i} \dfrac{\partial^2 f}{\partial x_i^2} \label{eq:5-36-3}\end{equation}

The operator $\displaystyle\sum_i \dfrac{\partial^2}{\partial x_i^2}$ on the right is called the Laplacian and is denoted by $\nabla^2$ or $\Delta$.

\begin{equation}\nabla \cdot (\nabla f) = \nabla^2 f = \Delta f \label{eq:5-36-4}\end{equation}

Proof

Take the dimension-independent expression from (36),

\begin{equation}\nabla^2 f = \displaystyle\sum_{i} \dfrac{\partial^2 f}{\partial x_i^2} \label{eq:5-37-1}\end{equation}

and substitute $(x_1, x_2, x_3) = (x, y, z)$.

\begin{equation}\nabla^2 f = \dfrac{\partial^2 f}{\partial x^2} + \dfrac{\partial^2 f}{\partial y^2} + \dfrac{\partial^2 f}{\partial z^2} \label{eq:5-37-2}\end{equation}

Proof

Write the $i$-th component of the curl with the Levi-Civita symbol $\varepsilon_{ijk}$:

\begin{equation}[\nabla \times \mathbf{V}]_i = \varepsilon_{ijk}\,\dfrac{\partial V_k}{\partial x_j} \label{eq:5-38-1}\end{equation}

Set $\mathbf{V} = \nabla f$, i.e. $V_k = \dfrac{\partial f}{\partial x_k}$:

\begin{equation}[\nabla \times (\nabla f)]_i = \varepsilon_{ijk}\,\dfrac{\partial}{\partial x_j}\!\left(\dfrac{\partial f}{\partial x_k}\right) = \varepsilon_{ijk}\,\dfrac{\partial^2 f}{\partial x_j \partial x_k} \label{eq:5-38-2}\end{equation}

Since $f \in C^2$, Schwarz's theorem (equality of mixed partials) gives

\begin{equation}\dfrac{\partial^2 f}{\partial x_j \partial x_k} = \dfrac{\partial^2 f}{\partial x_k \partial x_j} \label{eq:5-38-3}\end{equation}

so the second-derivative tensor is symmetric in $(j, k)$. The Levi-Civita symbol is antisymmetric in $(j, k)$:

\begin{equation}\varepsilon_{ijk} = -\varepsilon_{ikj} \label{eq:5-38-4}\end{equation}

The contraction of a symmetric tensor $S_{jk} = S_{kj}$ with an antisymmetric tensor $A_{jk} = -A_{kj}$ vanishes. Pairing the $(j, k)$ and $(k, j)$ terms in the sum,

\begin{equation}\varepsilon_{ijk}\,\dfrac{\partial^2 f}{\partial x_j \partial x_k} = \dfrac{1}{2}\!\left(\varepsilon_{ijk} + \varepsilon_{ikj}\right)\dfrac{\partial^2 f}{\partial x_j \partial x_k} = 0 \label{eq:5-38-5}\end{equation}

Every component vanishes, so

\begin{equation}\nabla \times (\nabla f) = \mathbf{0} \label{eq:5-38-6}\end{equation}

Proof

The $i$-th component of $\nabla \times \mathbf{A}$ is

\begin{equation}[\nabla \times \mathbf{A}]_i = \varepsilon_{ijk}\,\dfrac{\partial A_k}{\partial x_j} \label{eq:5-39-1}\end{equation}

Apply the definition of divergence $\nabla \cdot \mathbf{V} = \dfrac{\partial V_i}{\partial x_i}$ (summation convention).

\begin{equation}\nabla \cdot (\nabla \times \mathbf{A}) = \dfrac{\partial}{\partial x_i}\!\left(\varepsilon_{ijk}\,\dfrac{\partial A_k}{\partial x_j}\right) = \varepsilon_{ijk}\,\dfrac{\partial^2 A_k}{\partial x_i \partial x_j} \label{eq:5-39-2}\end{equation}

Since $\mathbf{A} \in C^2$, Schwarz's theorem gives

\begin{equation}\dfrac{\partial^2 A_k}{\partial x_i \partial x_j} = \dfrac{\partial^2 A_k}{\partial x_j \partial x_i} \label{eq:5-39-3}\end{equation}

so the second-derivative tensor is symmetric in $(i, j)$. The Levi-Civita symbol is antisymmetric in $(i, j)$ ($\varepsilon_{ijk} = -\varepsilon_{jik}$). By the same "symmetric × antisymmetric = 0" argument as in (38),

\begin{equation}\varepsilon_{ijk}\,\dfrac{\partial^2 A_k}{\partial x_i \partial x_j} = 0 \label{eq:5-39-4}\end{equation}

Therefore

\begin{equation}\nabla \cdot (\nabla \times \mathbf{A}) = 0 \label{eq:5-39-5}\end{equation}

Proof

Write the $i$-th component of the double curl in components:

\begin{equation}[\nabla \times (\nabla \times \mathbf{A})]_i = \varepsilon_{ijk}\,\dfrac{\partial}{\partial x_j}\,[\nabla \times \mathbf{A}]_k \label{eq:5-40-1}\end{equation}

Expand the inner curl with another Levi-Civita symbol:

\begin{equation}[\nabla \times \mathbf{A}]_k = \varepsilon_{klm}\,\dfrac{\partial A_m}{\partial x_l} \label{eq:5-40-2}\end{equation}

Substitute $\eqref{eq:5-40-2}$ into $\eqref{eq:5-40-1}$:

\begin{equation}[\nabla \times (\nabla \times \mathbf{A})]_i = \varepsilon_{ijk}\,\varepsilon_{klm}\,\dfrac{\partial^2 A_m}{\partial x_j \partial x_l} \label{eq:5-40-3}\end{equation}

Apply the contraction identity ($\varepsilon$-$\delta$ identity):

\begin{equation}\varepsilon_{ijk}\,\varepsilon_{klm} = \varepsilon_{kij}\,\varepsilon_{klm} = \delta_{il}\,\delta_{jm} - \delta_{im}\,\delta_{jl} \label{eq:5-40-4}\end{equation}

Substitute $\eqref{eq:5-40-4}$ into $\eqref{eq:5-40-3}$ and contract with the Kronecker deltas:

\begin{equation}[\nabla \times (\nabla \times \mathbf{A})]_i = \delta_{il}\,\delta_{jm}\,\dfrac{\partial^2 A_m}{\partial x_j \partial x_l} - \delta_{im}\,\delta_{jl}\,\dfrac{\partial^2 A_m}{\partial x_j \partial x_l} \label{eq:5-40-5}\end{equation}

The first term sets $l = i$, $m = j$; the second sets $m = i$, $l = j$.

\begin{equation}[\nabla \times (\nabla \times \mathbf{A})]_i = \dfrac{\partial^2 A_j}{\partial x_j \partial x_i} - \dfrac{\partial^2 A_i}{\partial x_j \partial x_j} \label{eq:5-40-6}\end{equation}

By Schwarz's theorem the order of partial derivatives in the first term may be exchanged, giving

\begin{equation}\dfrac{\partial^2 A_j}{\partial x_j \partial x_i} = \dfrac{\partial}{\partial x_i}\!\left(\dfrac{\partial A_j}{\partial x_j}\right) = \dfrac{\partial}{\partial x_i}(\nabla \cdot \mathbf{A}) = [\nabla(\nabla \cdot \mathbf{A})]_i \label{eq:5-40-7}\end{equation}

The second term is the component of the (Cartesian) Laplacian:

\begin{equation}\dfrac{\partial^2 A_i}{\partial x_j \partial x_j} = \displaystyle\sum_{j} \dfrac{\partial^2 A_i}{\partial x_j^2} = \nabla^2 A_i = [\nabla^2 \mathbf{A}]_i \label{eq:5-40-8}\end{equation}

Substituting $\eqref{eq:5-40-7}$ and $\eqref{eq:5-40-8}$ into $\eqref{eq:5-40-6}$:

\begin{equation}[\nabla \times (\nabla \times \mathbf{A})]_i = [\nabla(\nabla \cdot \mathbf{A})]_i - [\nabla^2 \mathbf{A}]_i \label{eq:5-40-9}\end{equation}

True for every $i$, so

\begin{equation}\nabla \times (\nabla \times \mathbf{A}) = \nabla(\nabla \cdot \mathbf{A}) - \nabla^2 \mathbf{A} \label{eq:5-40-10}\end{equation}

Proof

By (36) the Laplacian is the divergence of the gradient:

\begin{equation}\nabla^2(fg) = \nabla \cdot \bigl(\nabla(fg)\bigr) \label{eq:5-41-1}\end{equation}

The product rule for the gradient (identity (15), Leibniz rule) gives

\begin{equation}\nabla(fg) = f\,\nabla g + g\,\nabla f \label{eq:5-41-2}\end{equation}

Substitute $\eqref{eq:5-41-2}$ into $\eqref{eq:5-41-1}$:

\begin{equation}\nabla^2(fg) = \nabla \cdot (f\,\nabla g) + \nabla \cdot (g\,\nabla f) \label{eq:5-41-3}\end{equation}

Apply the divergence product rule (identity (23): $\nabla \cdot (h\mathbf{V}) = (\nabla h) \cdot \mathbf{V} + h\,(\nabla \cdot \mathbf{V})$) to each term.

\begin{equation}\nabla \cdot (f\,\nabla g) = (\nabla f) \cdot (\nabla g) + f\,\nabla \cdot (\nabla g) = \nabla f \cdot \nabla g + f\,\nabla^2 g \label{eq:5-41-4}\end{equation}

\begin{equation}\nabla \cdot (g\,\nabla f) = (\nabla g) \cdot (\nabla f) + g\,\nabla \cdot (\nabla f) = \nabla f \cdot \nabla g + g\,\nabla^2 f \label{eq:5-41-5}\end{equation}

Substitute $\eqref{eq:5-41-4}$ and $\eqref{eq:5-41-5}$ into $\eqref{eq:5-41-3}$:

\begin{equation}\nabla^2(fg) = f\,\nabla^2 g + 2(\nabla f \cdot \nabla g) + g\,\nabla^2 f \label{eq:5-41-6}\end{equation}

Proof

In Cartesian coordinates the basis vectors $\mathbf{e}_x, \mathbf{e}_y, \mathbf{e}_z$ are constant (independent of position), so any differential operator passes through the basis and acts only on the components.

The vector Laplacian is the quantity that appears as the second term on the right of (40); from $\eqref{eq:5-40-8}$ in that proof,

\begin{equation}[\nabla^2 \mathbf{A}]_i = \dfrac{\partial^2 A_i}{\partial x_j \partial x_j} = \displaystyle\sum_{j} \dfrac{\partial^2 A_i}{\partial x_j^2} = \nabla^2 A_i \label{eq:5-42-1}\end{equation}

For $i = 1, 2, 3$ (i.e. $i = x, y, z$),

\begin{equation}\nabla^2 \mathbf{A} = (\nabla^2 A_x,\ \nabla^2 A_y,\ \nabla^2 A_z) \label{eq:5-42-2}\end{equation}

Proof

By (42), in Cartesian coordinates the $i$-th component of $\nabla^2(f\mathbf{A})$ is $\nabla^2(fA_i)$:

\begin{equation}[\nabla^2(f\mathbf{A})]_i = \nabla^2(fA_i) \label{eq:5-43-1}\end{equation}

Apply (41) to the scalar fields $f$ and $A_i$:

\begin{equation}\nabla^2(fA_i) = (\nabla^2 f)\,A_i + 2(\nabla f \cdot \nabla A_i) + f\,\nabla^2 A_i \label{eq:5-43-2}\end{equation}

The term $\nabla f \cdot \nabla A_i$ is the $i$-th component of the directional-derivative operator $(\nabla f \cdot \nabla)$ applied to $\mathbf{A}$:

\begin{equation}\nabla f \cdot \nabla A_i = \displaystyle\sum_{j} \dfrac{\partial f}{\partial x_j}\,\dfrac{\partial A_i}{\partial x_j} = [(\nabla f \cdot \nabla)\mathbf{A}]_i \label{eq:5-43-3}\end{equation}

Substitute $\eqref{eq:5-43-3}$ into $\eqref{eq:5-43-2}$ and collect components:

\begin{equation}[\nabla^2(f\mathbf{A})]_i = (\nabla^2 f)\,A_i + 2[(\nabla f \cdot \nabla)\mathbf{A}]_i + f\,[\nabla^2 \mathbf{A}]_i \label{eq:5-43-4}\end{equation}

True for every $i$, hence

\begin{equation}\nabla^2(f\mathbf{A}) = (\nabla^2 f)\mathbf{A} + 2(\nabla f \cdot \nabla)\mathbf{A} + f\,\nabla^2\mathbf{A} \label{eq:5-43-5}\end{equation}

Proof

By the definition of the dot product,

\begin{equation}\mathbf{A} \cdot \mathbf{B} = \displaystyle\sum_{i} A_i\,B_i \label{eq:5-44-1}\end{equation}

By linearity of the Laplacian,

\begin{equation}\nabla^2(\mathbf{A} \cdot \mathbf{B}) = \displaystyle\sum_{i} \nabla^2(A_i\,B_i) \label{eq:5-44-2}\end{equation}

For each $i$, apply (41) to the scalar fields $A_i$ and $B_i$:

\begin{equation}\nabla^2(A_i\,B_i) = A_i\,\nabla^2 B_i + 2\,(\nabla A_i \cdot \nabla B_i) + B_i\,\nabla^2 A_i \label{eq:5-44-3}\end{equation}

Sum over $i$:

\begin{equation}\nabla^2(\mathbf{A} \cdot \mathbf{B}) = \displaystyle\sum_{i} A_i\,\nabla^2 B_i + 2\,\displaystyle\sum_{i} (\nabla A_i \cdot \nabla B_i) + \displaystyle\sum_{i} B_i\,\nabla^2 A_i \label{eq:5-44-4}\end{equation}

By (42), in Cartesian coordinates $[\nabla^2 \mathbf{B}]_i = \nabla^2 B_i$ and $[\nabla^2 \mathbf{A}]_i = \nabla^2 A_i$, so the first and third sums collapse into dot products:

\begin{equation}\displaystyle\sum_{i} A_i\,\nabla^2 B_i = \mathbf{A} \cdot \nabla^2 \mathbf{B},\quad \displaystyle\sum_{i} B_i\,\nabla^2 A_i = \mathbf{B} \cdot \nabla^2 \mathbf{A} \label{eq:5-44-5}\end{equation}

Substituting $\eqref{eq:5-44-5}$ into $\eqref{eq:5-44-4}$,

\begin{equation}\nabla^2(\mathbf{A} \cdot \mathbf{B}) = \mathbf{A} \cdot \nabla^2 \mathbf{B} + \mathbf{B} \cdot \nabla^2 \mathbf{A} + 2\,\displaystyle\sum_{i} (\nabla A_i \cdot \nabla B_i) \label{eq:5-44-6}\end{equation}

Proof

Apply the product rule for the divergence (identity (23)) to the vector field $f\,\nabla g$.

\begin{equation}\nabla \cdot (f\,\nabla g) = \nabla f \cdot \nabla g + f\,\nabla \cdot (\nabla g) = \nabla f \cdot \nabla g + f\,\nabla^2 g \label{eq:5-45-1}\end{equation}

The last equality uses (36), $\nabla \cdot (\nabla g) = \nabla^2 g$. Integrate $\eqref{eq:5-45-1}$ over $V$ and apply the divergence theorem to the left-hand side.

\begin{equation}\iiint_V \nabla \cdot (f\,\nabla g)\,dV = \oiint_S (f\,\nabla g) \cdot \mathbf{n}\,dS = \oiint_S f\,\dfrac{\partial g}{\partial n}\,dS \label{eq:5-45-2}\end{equation}

Here $\dfrac{\partial g}{\partial n} = \nabla g \cdot \mathbf{n}$ is the directional derivative along the outward normal. Equating with the integral of the right-hand side of $\eqref{eq:5-45-1}$ gives Green's first identity.

\begin{equation}\iiint_V f\,\nabla^2 g\,dV = -\iiint_V \nabla f \cdot \nabla g\,dV + \oiint_S f\,\dfrac{\partial g}{\partial n}\,dS \label{eq:5-45-3}\end{equation}

From here on assume in addition that $f \in C^2(\overline{V})$. Writing $\eqref{eq:5-45-3}$ again with $f$ and $g$ interchanged and subtracting, the term $\nabla f \cdot \nabla g$ cancels by symmetry and Green's second identity remains.

\begin{equation}\iiint_V \left(f\,\nabla^2 g - g\,\nabla^2 f\right)dV = \oiint_S \left(f\,\dfrac{\partial g}{\partial n} - g\,\dfrac{\partial f}{\partial n}\right)dS \label{eq:5-45-4}\end{equation}