The Gradient Map

Recall the definition of the gradient map $\nabla$ defined by

$\displaystyle \nabla := \sum_{i=1}^n \mathbf e_i \frac{\partial}{\partial x_i} : \mathcal K_1 \to \mathcal K_n,$

for sufficiently well-defined functions $\mathcal K_s \subseteq \mathcal F(\mathbb R^n, \mathbb R^s)$ . Functions in $\mathcal K_1$ are called scalar fields, while functions in $\mathcal K_n$ are called vector fields.

Definition 1. A map of the form $\phi : \mathbb R^n \to \mathbb R$ is called a scalar field. For such scalar fields, define the gradient of the scalar field using linearity:

$\displaystyle \nabla \phi = \sum_{i=1}^n \mathbf e_i \frac{\partial}{\partial x_i}(\phi) = \sum_{i=1}^n \frac{\partial \phi}{\partial x_i} \mathbf e_i.$

Problem 1. Verify the following properties for scalar fields $\psi,\phi$ , differentiable $f$ , and curves $\mathbf r$ .

$\nabla(\psi + \phi) = \nabla \psi + \nabla \phi$ ,
$\nabla(\psi \phi) = \phi \nabla \psi + \psi \nabla \phi$ ,
$\displaystyle \nabla \left( \frac{\psi}{\phi} \right) = \frac{ \phi \nabla \psi - \psi \nabla \phi }{\phi^2}$ ,
$\nabla (f \circ \phi) = (f' \circ \phi) \nabla \phi$ ,
$(\phi \circ \mathbf r)' = (\nabla \phi \circ \mathbf r) \circ \mathbf r'$ .

(Click for Solution)

Solution. The first result follows from linearity:

$\displaystyle \frac{\partial}{\partial x_i}( \psi + \phi) = \frac{\partial}{\partial x_i}(\psi) + \frac{\partial}{\partial x_i}(\phi),$

so that

$\begin{aligned} \nabla (\psi + \phi) &= \sum_{i=1}^n \mathbf e_i \frac{\partial}{\partial x_i}(\psi + \phi) \\ &= \sum_{i=1}^n \left[ \mathbf e_i \frac{\partial}{\partial x_i}(\psi) + \mathbf e_i \frac{\partial}{\partial x_i}(\phi) \right] \\ &= \sum_{i=1}^n \mathbf e_i \frac{\partial}{\partial x_i}(\psi) + \sum_{i=1}^n \mathbf e_i \frac{\partial}{\partial x_i}(\phi) \\ &= \nabla \psi + \nabla \phi. \end{aligned}$

The second result follows from the usual product rule:

$\displaystyle \frac{\partial}{\partial x_i}( \psi \phi) = \phi \frac{\partial}{\partial x_i}(\psi) + \psi \frac{\partial}{\partial x_i}(\phi) .$

The third result follows from the usual quotient rule:

$\displaystyle \frac{\partial}{\partial x_i} \left( \frac{ \psi }{ \phi } \right) = \frac 1{\phi^2} \left[ \phi \frac{\partial}{\partial x_i}(\psi) - \psi \frac{\partial}{\partial x_i}(\phi) \right].$

The fourth result follows from the usual chain rule:

$\displaystyle \frac{\partial}{\partial x_i} (f \circ \phi) = (f' \circ \phi) \frac{\partial }{\partial x_i}(\phi).$

The fifth result follows from the chain rule (Lemma 5 in this post).

Definition 2. A map of the form $\mathbf F : \mathbb R^m \to \mathbb R^n$ is called a vector field. For such vector fields, there exist scalar fields $f_1,\dots, f_n : \mathbb R^m \to \mathbb R$ such that

$\displaystyle \mathbf F = \sum_{i=1}^n f_i \mathbf e_i \equiv \begin{bmatrix} f_1 \\ \vdots \\ f_n \end{bmatrix}.$

Notice that a vector field reduces to a curve by setting $m = 1$ , and a scalar field by setting $n = 1$ . If $m = 2$ , we call $\mathbf F : \mathbb R^2 \to \mathbb R^n$ a surface.

Definition 3. Define the divergence of a vector field $\mathbf F = ( f_1,\dots, f_n )^{\mathrm T} : \mathbb R^n \to \mathbb R^n$ by

$\displaystyle \mathrm{div}(\mathbf F) := \sum_{i=1}^n \frac{\partial f_i}{\partial x_i}.$

By extending operator calculations using linearity,

$\begin{aligned} \nabla \cdot \mathbf F &= \left( \sum_{i=1}^n \mathbf e_i \frac{\partial}{\partial x_i} \right) \cdot \left( \sum_{j=1}^n f_j \mathbf e_j \right) \\ &= \sum_{i=1}^n \sum_{j=1}^n \left( \mathbf e_i \frac{\partial}{\partial x_i} \right) \cdot ( f_j \mathbf e_j ) \\ &= \sum_{i=1}^n \sum_{j=1}^n \frac{\partial f_j}{\partial x_i} ( \mathbf e_i \cdot \mathbf e_j ) \\ &= \sum_{i=1}^n \frac{\partial f_i}{\partial x_i} \\ &= \sum_{i=1}^n \frac{\partial }{\partial x_i}(f_i) = \mathrm{div}(\mathbf F). \end{aligned}$

Problem 2. Verify the following properties for the divergence of vector fields $\mathbf F, \mathbf G$ , scalar fields $\phi$ , and curves $\mathbf r$ .

$\nabla \cdot (\mathbf F + \mathbf G) = \nabla \cdot \mathbf F + \nabla \cdot \mathbf G$ ,
$\nabla \cdot (\phi \mathbf F) = \phi \nabla \cdot \mathbf F + (\nabla \phi) \cdot \mathbf F$ ,
$\displaystyle \nabla \cdot \left( \frac{ \mathbf F }{ \phi }\right) = \frac{ \phi \nabla \cdot \mathbf F - (\nabla \phi) \cdot \mathbf F }{ \phi^2 }$ ,
$\nabla \cdot (\mathbf r \circ \phi) = \nabla \phi \cdot (\mathbf r' \circ \phi)$ .

(Click for Solution)

Solution. We observe that $(\mathbf F+\mathbf G)_i = f_i + g_i$ . By the first result,

$\displaystyle \frac{\partial}{\partial x_i}( f_i+ g_i) = \frac{\partial f_i}{\partial x_i} + \frac{\partial g_i}{\partial x_i}.$

Hence,

$\begin{aligned} \nabla \cdot (\mathbf F + \mathbf G) &= \sum_{i=1}^n \frac{\partial}{\partial x_i} (( \mathbf F + \mathbf G )_i) \\ &= \sum_{i=1}^n \frac{\partial }{\partial x_i} (f_i + g_i) \\ &= \sum_{i=1}^n \left[ \frac{\partial }{\partial x_i} (f_i) + \frac{\partial }{\partial x_i} (g_i) \right] \\ &= \sum_{i=1}^n \frac{\partial }{\partial x_i} (f_i) + \sum_{i=1}^n \frac{\partial }{\partial x_i} (g_i) \\ &= \nabla \cdot \mathbf F + \nabla \cdot \mathbf G. \end{aligned}$

We observe that $(\phi \mathbf F)_i = \phi f_i$ . The product rule yields

$\displaystyle \frac{\partial}{\partial x_i} (\phi f_i) = \phi \frac{\partial}{\partial x_i} (f_i) + f_i \frac{\partial}{\partial x_i} (\phi).$

Summing over $i$ yields

$\begin{aligned} \sum_{i=1}^n \frac{\partial}{\partial x_i} (\phi f_i) &= \phi \sum_{i=1}^n \frac{\partial}{\partial x_i} (f_i) + \sum_{i=1}^n f_i \frac{\partial}{\partial x_i} (\phi) \\ &= \phi \nabla \cdot \mathbf F + \mathbf F \cdot \nabla \phi \\ &= \phi \nabla \cdot \mathbf F + (\nabla \phi) \cdot \mathbf F. \end{aligned}$

Similarly, the quotient rule yields

$\displaystyle \frac{\partial}{\partial x_i} \left( \frac{ f_i }{ \phi } \right) = \frac 1{\phi^2} \left[ \phi \frac{\partial}{\partial x_i}(f_i) - f_i \frac{\partial}{\partial x_i}(\phi) \right].$

Finally, given the curve $\mathbf r(t) = (r_1(t),\dots, r_n(t))^{\mathrm T}$ , the chain rule yields

$\displaystyle \frac{ \partial }{ \partial x_i } (\mathbf r \circ \phi)_i = \frac{ \partial }{ \partial x_i } (r_i \circ \phi) = ( r_i' \circ \phi ) \frac{ \partial \phi }{ \partial x_i }.$

Summing over $i$ yields

$\begin{aligned} \nabla \cdot (\mathbf r \circ \phi) &= \sum_{i=1}^n \frac{\partial}{\partial x_i} (\mathbf r \circ \phi)_i \\ &= \sum_{i=1}^n ( r_i' \circ \phi ) \frac{ \partial \phi }{ \partial x_i } \\ &= \nabla \phi \cdot (\mathbf r' \circ \phi). \end{aligned}$

Definition 4. Define the Laplacian $\Delta$ of a scalar field $\phi : \mathbb R^n \to \mathbb R$ by

$\Delta \phi := \nabla \cdot (\nabla \phi) = (\nabla \cdot \nabla) \phi = \nabla^2 \phi.$

Hence, we can use the abbreviation $\Delta = \nabla^2 = \nabla \cdot \nabla$ .

Problem 3. Verify the following properties involving Laplacians, given scalar fields $\psi, \phi$ .

$\Delta (\psi \phi) = \psi \Delta \phi + \phi \Delta \psi + 2 \nabla \psi \cdot \nabla \phi$ ,
$\displaystyle \Delta \left( \frac{\psi}{\phi} \right) = \frac{ \psi \Delta \phi - \phi \Delta \psi - 2 \phi \nabla ( \psi / \phi ) \cdot \nabla \phi }{ \phi^2 }.$

(Click for Solution)

Solution. Using Problem 1 and Problem 2,

$\begin{aligned} \Delta (\psi \phi) &= \nabla \cdot (\nabla(\psi \phi)) \\ &= \nabla \cdot (\phi \nabla \psi + \psi \nabla \phi) \\ &= \nabla \cdot( \phi \nabla \psi) + \nabla \cdot(\psi \nabla \phi) \\ &= \phi \nabla \cdot (\nabla \psi) + \nabla \phi \cdot \nabla \psi + \psi \nabla \cdot (\nabla \phi) + \nabla \psi \cdot \nabla \phi \\ &= \phi \Delta \psi + \psi \Delta \phi + 2\nabla \phi \cdot \nabla \psi. \end{aligned}$

Replacing $\psi$ with $\psi/\phi$ in the first result,

$\displaystyle \Delta \phi = \phi \Delta \left( \frac{\psi}{\phi} \right) + \frac{\psi}{\phi} \Delta \phi + 2 \nabla(\psi/\phi) \cdot \nabla \phi.$

The result follows by making $\Delta(\psi/\phi)$ the subject.

Definition 4. Define the curl of a 3-dimensional vector field

$\mathbf F = ( f_1, f_2, f_3)^{\mathrm T} : \mathbb R^3 \to \mathbb R^3$

$\displaystyle \mathrm{curl}(\mathbf F) := \det(\nabla, \mathbf F, \cdot) \equiv \nabla \times \mathbf F$

using similar notational simplification.

Problem 4. Verify the following properties for the curl of 3-dimensional vector fields $\mathbf F, \mathbf G$ and scalar fields $\phi$ .

$\nabla \times (\mathbf F + \mathbf G) = \nabla \times \mathbf F + \nabla \times \mathbf G$ ,
$\nabla \times (\phi \mathbf F) = \phi \nabla \times \mathbf F + ( \nabla \phi ) \times \mathbf F$ ,
$\displaystyle \nabla \times \left( \frac{ \mathbf F }{ \phi }\right) = \frac{ \phi \nabla \times \mathbf F - (\nabla \phi) \times \mathbf F }{ \phi^2 }$ ,
$\nabla \times (\mathbf r \circ \phi) = \nabla \phi \times (\mathbf r' \circ \phi)$ ,
$\nabla \cdot (\nabla \times \mathbf F) = 0$ ,
$\nabla \times (\nabla \phi) = \mathbf 0$ .

(Click for Solution)

Solution. For the first result, use the multi-linearity of $\det$ : for any $\mathbf v \in \mathbb R^3$ ,

$\det(\nabla, \mathbf F + \mathbf G, \mathbf v) = \det(\nabla, \mathbf F, \mathbf v) + \det(\nabla, \mathbf G, \mathbf v) .$

Therefore,

$\begin{aligned}\nabla \times (\mathbf F + \mathbf G) &= \det(\nabla, \mathbf F + \mathbf G, \cdot) \\ &= \det(\nabla, \mathbf F, \cdot) + \det(\nabla, \mathbf G, \cdot) \\ &= \nabla \times \mathbf F + \nabla \times \mathbf G.\end{aligned}$

By definition of the cross product,

$\displaystyle \nabla \times \mathbf F = \left( \frac{\partial f_3}{\partial x_2} - \frac{\partial f_2}{\partial x_3} \right)\mathbf e_1 + \left( \frac{\partial f_1}{\partial x_3} - \frac{\partial f_3}{\partial x_1} \right)\mathbf e_2 + \left( \frac{\partial f_2}{\partial x_1} - \frac{\partial f_1}{\partial x_2} \right)\mathbf e_3.$

Denoting $\displaystyle g_i = \frac{\partial}{\partial x_i}(g)$ ,

$\displaystyle \nabla \times \mathbf F = \begin{bmatrix} (f_3)_2 - (f_2)_3 \\ (f_1)_3 - (f_3)_1 \\ (f_2)_1 - (f_1)_2 \end{bmatrix}.$

The second result then follows by $(\phi \mathbf F)_i = \phi f_i$ and using the vanilla product rule. The third result follows from the vanilla quotient rule. The fourth result follows from the vanilla chain rule. The fifth result follows from Clairaut’s theorem, which states that for sufficiently nice $g$ , $(g_i)_j = (g_j)_i$ . Denoting $g_{ij} := (g_i)_j$ ,

$\begin{aligned} \nabla \cdot (\nabla \times \mathbf F) &= \frac{\partial}{\partial x_1} ( (f_3)_2 - (f_2)_3 ) + \frac{\partial}{\partial x_2} ( (f_1)_3 - (f_3)_1 )+ \frac{\partial}{\partial x_3} ( (f_2)_1 - (f_1)_2 ) \\ &= (f_3)_{21} - (f_2)_{31} + (f_1)_{32} - (f_3)_{12} + (f_2)_{13} - (f_1)_{23} \\ &= (f_3)_{12} - (f_2)_{13} + (f_1)_{23} - (f_3)_{12} + (f_2)_{13} - (f_1)_{23} = 0. \end{aligned}$

The sixth result follows from $(\nabla \phi)_i = \displaystyle \frac{ \partial }{\partial x_i}(\phi) \equiv \phi_i$ and Clairaut’s theorem:

$\displaystyle \nabla \times \mathbf F = \begin{bmatrix} (\phi_3)_2 - (\phi_2)_3 \\ (\phi_1)_3 - (\phi_3)_1 \\ (\phi_2)_1 - (\phi_1)_2 \end{bmatrix} = \begin{bmatrix} (\phi_3)_2 - (\phi_3)_2 \\ (\phi_1)_3 - (\phi_1)_3 \\ (\phi_1)_2 - (\phi_1)_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} = \mathbf 0.$

Remark 1. Using more brute-force, we can verify the following combined gradient properties in 3-dimensional space for vector fields $\mathbf F, \mathbf G$ and scalar fields $\psi, \phi$ .

$\nabla \cdot (\mathbf F \times \mathbf G) = (\nabla \times \mathbf F) \cdot \mathbf G - (\nabla \times \mathbf G) \cdot \mathbf F$ ,
$\nabla \times (\psi \nabla \phi) = \nabla \psi \times \nabla \phi$ ,
$\nabla \times (\mathbf F \times \mathbf G) = \mathbf F (\nabla \mathbf G) - \mathbf G (\nabla \mathbf F) + (\mathbf F \cdot \nabla) \mathbf G - (\mathbf F \cdot \nabla) \mathbf G$ ,
$\Delta (\nabla \times \mathbf F) = \nabla \times \Delta \mathbf F$ .