Multivariable Differential Operators
First, some preliminary definitions.
Let $P$ be any point in a domain of definition. Then we define a vector function $\vec{v}$ whose values are vectors, that is,
$$ \vec{v} = \vec{v}(P) = [v_1(P), v_2(P), v_3(P)] $$
that depends on points $P$ in space. A vector function depends only on the point $P,$ not on the coordinate system chosen to represent its components.
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We say that a vector function defines a vector field in a domain of definition.
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Let $P$ be any point in a domain of definition. Then we define a scalar function $f,$ whose values are scalars, that is,
$$ f = f(p) $$
that depends on $P.$
A scalar function depends only on the point $P,$ not on the coordinate system chosen to represent it.
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Let $\vec{f}$ be a function that maps an open set $E \subset R^n$ into $R^m.$ Let $\{\vec{e}_1, \dots, \vec{e}_n\}$ and $\{\vec{u}_1, \dots, \vec{u}_n\}$ be the standard bases of $R^n$ and $R^m.$ The components of $\vec{f}$ are the real functions $f_1, \dots, f_m$ defined by
$$ \vec{f}(\vec{x}) = \sum_{i=1}^m f_i(\vec{x})\vec{u}_i \quad (\vec{x} \in E), $$
or, equivalently, by $f_i(\vec{x}) = \vec{f}(\vec{x}) \cdot \vec{u}_i, 1 \leq i \leq m. $
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- proof-of-theorem-19
- remark-45
- Normal Derivative
- directional-derivative-is-inner-product-of-vector-and-grad
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Using the setup from the definition of component, for $\vec{x} \in E, 1 \leq i \leq m, 1 \leq j \leq n,$ we define
$$ (D_j f_i)(\vec{x}) = \lim_{t \to 0} \frac{f_i(\vec{x} + t \vec{e}_j) - f_i(\vec{x})}{t}, $$
provided the limit exists. Writing $f_i(x_1, \dots, x_n)$ in place of $f_i(\vec{x}),$ we see that $D_j f_i$ is the derivative of $f_i$ with respect to $x_j,$ keeping the other variables fixed. The notation
$$ \frac{\partial f_i}{\partial x_j} $$
is therefore often used in place of $D_j f_i,$ and $D_j f_i$ is called a partial derivative.
For a space curve given parametrically by $\vec{r}(t),$ the tangent vector (or specifically, the unit tangent vector) at the point $\vec{r}(t)$ is the unit vector defined by:
$$ \vec{T}(t) = \frac{\vec{r}'(t)}{||\vec{r}'(t)||} $$
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Differentiation
If $f$ is a real function with domain of definition $(a, b) \subset R^1$ and if $x \in (a,b),$ then $f'(x)$ is usually defined to be the real number
$$ lim_{h \to 0} \frac{(x+h) - f(x)}{h}, $$
provided this limit exists. Thus,
$$ f(x + h) - f(x) = f'(x)h + r(h) \tag{8} $$
where the "remainder" $r(h)$ is small, that is,
$$ \lim_{h \to 0} = \frac{r(h)}{h} = 0. $$
Note that (8) expresses the difference $f(x+h) - f(x)$ as the sum of the linear function that takes $h$ to $f'(x)h,$ plus a small remainder. We can therefore view the derivative of $f$ at $x$ as the linear operator on $R^1$ that takes $h$ to $f'(x)h.$
On linearity: every real number $\alpha$ gives rise to a linear operator on $R^1:$
$$ L_\alpha{x} = \alpha x. $$
Conversely, every linear function from $R^1$ to $R^1$ is multiplication by some real number. Thus, we have a $1-1$ correspondence between $R^1$ and $L(R^1)$ (the set of all linear transformations on $R^1$.)
Now, consider a function $\vec{f}$ that maps $(a, b) \subset R^1$ into $R^m.$ Then $\vec{f}'(x)$ is defined to be the vector $y \in R^m$ (if one exists) for which
$$ \lim_{h \to 0} \left ( \frac{\vec{f}(x+h) - \vec{f}(x)}{h} - \vec{y} \right ) = 0. $$
This can be rewritten as
$$ \vec{f}(x + h) - \vec{f}(x) = h \vec{y} + \vec{r}(h), $$
where $\vec{r}(h)/h \to \vec{0}$ as $h \to 0.$ Again, the right side is a linear function of $h.$ Every $\vec{y} \in R^m$ gives a linear transformation of $R^1$ into $R^m:$
$$L_\vec{y}{h} = h \vec{y}, $$
which gives us an identification of $R^m$ with $L(R^1, R^m)$ (the set of all linear transformations from $R^1$ to $R^m$,) allowing us to say $\vec{f}'(x) \in L(R^1, R^m)$.
Therefore, if $\vec{f}$ is a differentiable function of $(a, b) \subset R^1$ into $R^m,$ and if $x \in (a, b),$ then $\vec{f}'(x)$ is the linear transformation of $R^1$ into $R^m$ that satisfies
$$ \lim_{h \to 0} \frac{\vec{f}(x + h) - \vec{f}(x) - \vec{f}'(x) h}{h} = \vec{0}, $$
or, equivalently,
$$ \lim_{h \to 0} \frac{|\vec{f}(x + h) - \vec{f}(x) - \vec{f}'(x) h|}{|h|} = 0. $$
We can now take on the general $R^n \to R^m$ case.
Suppose $E$ is an open set in $R^n,$ $\vec{f} : E \to R^m,$ and $\vec{x} \in E.$ If there exists a linear transformation $A : R^n \to R^m$ such that
$$ \lim_{\vec{h} \to \vec{0}} \frac{|\vec{f}(\vec{x} + \vec{h}) - \vec{f}(\vec{x}) - A\vec{h}|}{|\vec{h}|} = 0, \tag{14} $$
then we say that $\vec{f}$ is differentiable at $\vec{x}$ and we write
$$ \vec{f}'(\vec{x}) = A. $$
If $\vec{f}$ is differentiable at every $\vec{x} \in E,$ we say that $\vec{f}$ is differentiable in $E.$
Note that in (14), $\vec{h} \in R^n.$ If $|\vec{h}|$ is small enough, then $\vec{x} + \vec{h} \in E,$ because $E$ is open. Therefore, $\vec{f}(\vec{x} + \vec{h})$ is defined, $\vec{f}(\vec{x} + \vec{h}) \in R^m,$ and since $A \in L(R^n, R^m),$ $A \vec{h} \in R^m.$ Therefore,
$$ \vec{f}(\vec{x} + \vec{h}) - \vec{f}(\vec{x}) - A \vec{h} \in R^m. $$
The norm in the numerator of (14) is that of $R^m,$ while the norm in the denominator is the $R^n$-norm.
We can also rewrite (14) as
$$ \vec{f}(\vec{x} + \vec{h}) - \vec{f}(\vec{x}) = \vec{f}'(\vec{x})\vec{h} + \vec{r}(\vec{h}), \tag{17} $$
where the remainder $\vec{r}(\vec{h})$ satisfies
$$ \lim_{\vec{h} \to \vec{0}} \frac{|\vec{r}(\vec{h})|}{|\vec{h}|} = 0. $$
This means that for a fixed $\vec{x}$ and a small $\vec{h},$ the left side of (17) is approximately equal to $\vec{f}'(\vec{x})\vec{h},$ that is, to the value of a linear transformation applied to $\vec{h}.$
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The derivative defined above in differentiable is called the total derivative of $\vec{f}$ at $\vec{f},$ or the differential of $\vec{f}$ at $\vec{x}.$
Suppose $E$ is an open set in $R^n,$ $\vec{f} : E \to R^m,$ and $\vec{x} \in E,$ and that $A_1$ and $A_2$ are total derivatives of $\vec{f}$ at $\vec{x}.$ Then, $A_1 = A_2.$
A differentiable function $\vec{f}$ of an open set $E \subset R^n$ into $R^m$ is said to be continuously differentiable in $E$ if $\vec{f}'$ is a continuous function of $E$ into $L(R^n, R^m).$ More explicitly, it is required that to every $\vec{x} \in E$ and to every $\epsilon > 0$ corresponds a $\delta > 0$ such that
$$ ||\vec{f}'(\vec{y}) - \vec{f}'(\vec{x})|| < \epsilon $$
if $\vec{y} \in E$ and $|\vec{x} - \vec{y}| < \delta.$
If this is the case, we also say that $\vec{f}$ is a $\mathscr{C}'$-mapping or that $\vec{f} \in \mathscr{C}'(E).$
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These are some differential operators that apply to multivariable and/or vector valued functions.
Vector Differential Operator
In Euclidean space $\mathbb{R}^n$ with coordinates $(x_1, \dots, x_n)$ and standard basis $(\vec{e}_1, \dots, \vec{e}_n),$ del is a vector operator whose $x_1, \dots, x_n$ components are the partial derivative operators $\frac{\partial}{\partial x_1}, \dots, \frac{\partial}{\partial x_n}; $ that is
$$ \nabla = \sum_{i = 1}^n \vec{e}_i \frac{\partial}{\partial x_i} = \left ( \frac{\partial}{\partial x_1}, \dots, \frac{\partial}{\partial x_n} \right ). $$
Gradient
Given a scalar function $f : \mathbb{R}^n \to \mathbb{R}$, the gradient of $f$, denoted as $\nabla f$, is defined as the vector of its partial derivatives. Specifically, for a function $f(x_1, x_2, \cdots, x_n)$, the gradient is given by
$$ \nabla f = \begin{bmatrix} \frac{\partial f}{\partial x_1} \\ \frac{\partial f}{\partial x_2} \\\ \vdots \\\ \frac{\partial f}{\partial x_n} \end{bmatrix} $$
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We can write the gradient as either a row vector or a column vector. Different texts use different conventions. I'll show it here as a row vector, while we used a column vector above, just to mix it up.
In three-dimensional Euclidean space, the gradient of a function $f,$ if it exists, is given by
$$ \grad{f} = \nabla f = \frac{\partial f}{\partial x} \vec{i} + \frac{\partial f}{\partial y} \vec{j} + \frac{\partial f}{\partial z} \vec{k} = \left [ \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right ] $$
Directional Derivative
Using the setup from the definition of component, let us fix an $\vec{x} \in E$ (with $E \subset R^n$,) and let $\vec{u} \in R^n$ be a unit vector. Then,
$$ \lim_{t \to 0} \frac{f(\vec{x} + t \vec{u}) - f(\vec{x})}{t} = (\nabla f)(\vec{x}) \cdot \vec{u} $$
is called the directional derivative of $f$ at $\vec{x}$ in the direction of the unit vector $u$ and is denoted as $D_{\vec{u}}f(\vec{x}).$
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The directional derivative of $f$ in the direction of a unit vector $\vec{u}$ is the inner product of $\vec{u}$ and $\grad{f},$ that is,
$$ D_{\vec{u}} = \vec{u} \cdot \grad{f}. $$
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Let $f(P) = f(x,y,z)$ be a scalar function having continuous first partial derivatives in some domain $B$ in space. Then, $\grad{f}$ exists in $B$ and is a vector, that is, its length and direction are independent of the particular choice of Cartesian coordinates. If $\grad{f(P)} \neq \vec{0}$ at some point $P,$ it has the direction of maximum increase of $f$ at $P.$
The directional derivative of $f$ in the direction of some unit vector $\vec{u}$ is
$$ D_\vec{u} f = \vec{u} \cdot \grad{f} = |\vec{u}| |\grad{f}| \cos{\theta} \tag{a} $$
where $\theta$ is the angle between $\vec{u}$ and $\grad{f}$ (see The @directional-derivative of $f$ in the @direction... and The @dot-product of $\vec{u}$ and $\vec{v}$ is...). Note that $f$ is a scalar function, as is the directional derivative of $f.$ Now, $cos{\theta}$ has its maximum value of $1$ whenever $\theta = 0,$ and since $\vec{u}$ is a unit vector with magnitude of 1, (a) simplifies to
$$ D_{\vec{u}} f = |\grad{f}|, $$
which tells us that that direction and magnitude of $\grad{f}$ are independent of the coordinate system chosen. Now, since $\theta = 0$ if and only if $\vec{b}$ and $\grad{f}$ are parallel, $\grad{f}$ is the direction of maximum increase of $f$ at $P,$ assuming $\grad{f} \neq 0$ at $P.$
$\square$Let $S$ be a surface represented by $f(x, y, z) = c,$ where $c$ is constant and $f$ is differentiable. Such a surface is called a level surface of $f,$ and for different $c,$ we get different level surfaces.
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If $S$ is a level surface of a function $f,$ and $P$ is a point of $S,$ then the set of all tangent vectors of all curves passing through $P$ will generally form a plane, called the tangent plane of $S$ at $P.$
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Given a tangent plane to a surface $S$ at $P,$ the normal to this plane (the straight line through $P$ perpendicular to the tangent plane) is called the surface normal to $S$ at $P.$
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A vector in the direction of the surface normal of a surface $S$ at point $P$ is called a surface normal vector of $S$ at $P.$
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Let $f$ be a differentiable scalar function in space. Let $f(x,y,z) = c$ (with $c$ constant) represent a surface $S.$ Then, if the gradient of $f$ at a point $P$ of $S$ is not the zero vector, it is a surface normal vector of $S$ at $P.$
Any curve $C$ lying in $S$ can be parameterized as $\vec{r} = [x(t), y(t), z(t)]$ such that
$$ f(x(t), y(t), z(t)) = c. \tag{a} $$
Now, if we differentiate (a) with respect to $t,$ we get
$$ \frac{df}{dt} = \frac{\partial f}{\partial x} x' + \frac{\partial f}{\partial y} y' + \frac{\partial f}{\partial z} z' = (\grad{f}) \cdot \vec{r}' = 0. $$
Therefore, $\grad{f}$ is orthogonal to all the vectors $\vec{r}'$ in the tangent plane of $S$ at $P,$ and is therefore a surface normal vector of $S$ at $P.$
$\square$If $S$ is given by $F(x,y,z) = 0,$ then at $P = (x_0, y_0, z_0),$ the tangent plane is
$$ \nabla F(P) \cdot \langle x - x_0, y - y_0, z - z_0 \rangle = 0. $$
A potential function is a scalar function whose gradient is a vector field. They allow representing certain vector fields in a simpler, more fundamental form.
In other words, given a vector field $\vec{F},$ a potential function $\phi$ is a scalar function such that
$$ \vec{F} = - \grad{\phi} \quad \text{or} \quad \vec{F} = \grad{\phi}. $$
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Normal Derivative
The normal derivative is the directional derivative in the direction of the @normal-vector.
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The normal derivative tells us how the value of a function changes as we move in the direction normal (orthogonal) to a curve or surface.
Divergence
For a vector field $\vec{F} = (F_1, F_2, F_3)$ defined in three-dimensional space $\mathbb{R}^3$, the divergence is defined as
$$ \div \vec{F} = \nabla \cdot \vec{F} = \frac{\partial F_1}{\partial x} + \frac{\partial F_2}{\partial y} + \frac{\partial F_3}{\partial z}. $$
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The divergence of a vector field quantifies the extent to which the vector field behaves as a source or a sink at a given point.
divergence gives a scalar value for each point, i.e., it is a scalar field. A positive value indicates a net flow away from the point, while a negative value indicates a net flow towards a point.
An important theorem related to divergence is the Divergence Theorem (also known as Gauss's theorem) which connects the flux of a vector field through a closed surface to the divergence of the field inside the volume bounded by the surface:
$$ \iiint_V (\nabla \cdot \vec{F}) dV = \oint_s \vec{F} \cdot d\vec{S} $$
Let $\vec{v}$ be the velocity vector of of the motion of particles in a fluid. If $\div{\vec{v}} = 0,$ then the fluid has constant density and is said to be incompressible.
Curl
For a vector field $\vec{F} = (F_1, F_2, F_3)$ defined in three-dimensional space $\mathbb{R}^3$, with each component function $F_i$ depending on the variables $x$, $y$, and $z$, the curl of $\vec{F}$ is defined as
$$ \curl \vec{F} = \nabla \times \vec{F} = \left ( \frac{\partial F_3}{\partial y} - \frac{\partial F_2}{\partial z} \right ) \mathbf{\vec{i}} + \left ( \frac{\partial F_1}{\partial z} - \frac{\partial F_3}{\partial x} \right ) \mathbf{\vec{j}} + \left ( \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y} \right ) \mathbf{\vec{k}} $$
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The curl at a point in the field is represented by a vector whose length and direction denote the magnitude and axis of the maximum circulation. Circulation is the line integral of a vector field around a closed curve.
More intuitively, curl measures the rotation of the vector field at a given point.
Curl can also be expressed as the determinant of a 3x3 matrix involving the unit vectors, partial derivatives, and the components of the vector field:
$$ \nabla \times \vec{F} = \begin{vmatrix} \mathbf{\vec{i}} & \mathbf{\vec{j}} & \mathbf{\vec{k}} \\\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\\ F_1 & F_2 & F_3 \end{vmatrix} $$
If the curl of a vector field is $\vec{0},$ i.e. if $\curl{\vec{v}} = 0,$ the field is said to be irrotational.
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Laplacian
The Laplace operator is a second-order differential operator in the $n$-dimensional Euclidean space, defined as the divergence $(\nabla \cdot)$ of the gradient $(\nabla f$). Thus, if $f$ is a twice differentiable real-valued function, then the Laplacian of $f$ is the real-valued function defined by
$$ \Delta f = \nabla^2 f = \nabla \cdot \nabla f. $$
The Laplacian of $f$ is the sum of all the unmixed second partial derivatives in the Cartesian coordinates $x_i$:
$$ \nabla^2 f = \sum_{i=1}^n \frac{\partial^2 f}{\partial x_i^2}. $$
In two dimensions, using Cartesian coordinates, the Laplace operator is given by
$$ \nabla^2 f = \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} $$
and in three dimensions by
$$ \nabla^2 f = \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} + \frac{\partial^2 f}{\partial z^2} $$
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Solutions to Laplace's equation $\nabla^2 f = 0$ are called [[harmonic functions|analysis/complex-analysis/module-06-harmonic]].
The force of attraction
$$ \vec{p} = - \frac{c}{r^3} \vec{r} = -c \left [ \frac{x - x_0}{r^3}, \frac{y - y_0}{r^3}, \frac{z - z_0}{r^3} \right ] $$
between two particles at points $P_0 = (x_0, y_0, z_0)$ and $P = (x, y, z)$ (as given by Newton's law of gravitation) has the potential function $f(x, y, z) = c/r,$ where $r > 0$ is the distance between $P_0$ and $P.$
Thus, $\vec{p} = \grad{f} = \grad(c/r).$ This potential function $f$ is a solution of Laplace's Equation
$$ \nabla^2 f = \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} + \frac{\partial^2 f}{\partial z^2} = 0, $$
that is, $f$ has a laplacian of $0.$
Hessian
Suppose $f ~ : ~ \mathbb{R}^n \to \mathbb{R}$ is a function taking as input a vector $\vec{x} \in \mathbb{R}^n$ and outputting a scalar $f(\vec{x}) \in \mathbb{R}.$ If all second-order partial derivatives of $f$ exist, then the Hessian matrix $\vec{H}$ of $f$ is a square $n \times n$ @matrix, usually defined and arranged as
$$ \mathbf H_f= \begin{bmatrix} \dfrac{\partial^2 f}{\partial x_1^2} & \dfrac{\partial^2 f}{\partial x_1\,\partial x_2} & \cdots & \dfrac{\partial^2 f}{\partial x_1\,\partial x_n} \\[2.2ex] \dfrac{\partial^2 f}{\partial x_2\,\partial x_1} & \dfrac{\partial^2 f}{\partial x_2^2} & \cdots & \dfrac{\partial^2 f}{\partial x_2\,\partial x_n} \\[2.2ex] \vdots & \vdots & \ddots & \vdots \\[2.2ex] \dfrac{\partial^2 f}{\partial x_n\,\partial x_1} & \dfrac{\partial^2 f}{\partial x_n\,\partial x_2} & \cdots & \dfrac{\partial^2 f}{\partial x_n^2} \end{bmatrix}. $$
That is, the @entry of the $i$th row and the $j$th column is
$$ (\vec{H}_f)_{i,j} = \frac{\partial^2 f}{\partial x_i \partial x_j}. $$
For a function $f ~ : ~ \mathbb{R}^3 \to \mathbb{R},$ this is
$$ \vec{H} = \begin{bmatrix} \dfrac{\partial^2 f}{\partial x^2} & \dfrac{\partial^2 f}{\partial x \partial y} & \dfrac{\partial^2 f}{\partial x \partial z}\\ \dfrac{\partial^2 f}{\partial y \partial x} & \dfrac{\partial^2 f}{\partial y^2} & \dfrac{\partial^2 f}{\partial y \partial z}\\ \dfrac{\partial^2 f}{\partial z \partial x} & \dfrac{\partial^2 f}{\partial z \partial y} & \dfrac{\partial^2 f}{\partial z^2}\\ \end{bmatrix}. $$
Note that the laplacian of $f$ is equal to the @trace of the hessian of $f.$ In this way, the laplacian is to the hessian what divergence is to the jacobian.
Jacobian
Let $\vec{f} ~ : ~ \mathbb{R}^n \to \mathbb{R}^m$ be a function such that each of its first-order partial derivatives exists on $\mathbb{R}^n.$ This function takes a point $\vec{x} = (x_1, \dots, x_n) \in \mathbb{R}^n$ as input and produces the vector $\vec{f}(\vec{x}) = (f_1(\vec{x}), \dots, f_m(\vec{x})) \in \mathbb{R}^m$ as output. Then the Jacobian matrix of $\vec{f},$ denoted $\vec{J}_{\vec{f}},$ is the $m \times n$ @matrix whose $(i,j)$ @entry is $\frac{\partial f_i}{\partial x_j};$ explicitly
$$ \begin{bmatrix} \dfrac{\partial \mathbf{f}}{\partial x_1} & \cdots & \dfrac{\partial \mathbf{f}}{\partial x_n} \end{bmatrix} = \begin{bmatrix} \nabla^{\mathsf{T}} f_1 \\ \vdots \\ \nabla^{\mathsf{T}} f_m \end{bmatrix} = \begin{bmatrix} \dfrac{\partial f_1}{\partial x_1} & \cdots & \dfrac{\partial f_1}{\partial x_n}\\ \vdots & \ddots & \vdots\\ \dfrac{\partial f_m}{\partial x_1} & \cdots & \dfrac{\partial f_m}{\partial x_n} \end{bmatrix}, $$
where $\nabla^{\mathsf{T}} f_i$ is the @transpose (@row-vector) of the gradient of the $i$-th component.
$$ \vec{F} : \mathbb{R}^n \to \mathbb{R}^m, $$
the jacobian $\vec{J}(\vec{x})$ is the @linear-map that best @approximates $\vec{F}$ near $\vec{x}:$
$$ \vec{F}(\vec{x_0} + \vec{\Delta x}) \approx \vec{F(x_0)} + \vec{J(x_0)} \vec{\Delta x}. $$
At a point $x_0 \in \mathbb{R}^n,$ the directional derivative of $F$ in the direction of $\vec{v} \in \mathbb{R}^n$ is
$$ D_{\vec{v}} \vec{F(x_0)} = \vec{J(x_0)} \vec{v}. $$
Jacobian Related to Gradient
Given a vector field $F:$
$$ \vec{F} = \begin{bmatrix} F_1(x,y,z) \\ F_2(x,y,z) \\ F_3(x,y,z) \\ \end{bmatrix} : \mathbb{R}^3 \to \mathbb{R}^3, $$
each component $F_i$ is itself a scalar field, and so we can take the gradient of each one:
$$ \nabla F_1 = \begin{bmatrix} \dfrac{\partial F_1}{\partial x} \\ \dfrac{\partial F_1}{\partial y} \\ \dfrac{\partial F_1}{\partial z} \\ \end{bmatrix}, \quad \nabla F_2 = \begin{bmatrix} \dfrac{\partial F_2}{\partial x} \\ \dfrac{\partial F_2}{\partial y} \\ \dfrac{\partial F_2}{\partial z} \\ \end{bmatrix}, \quad \nabla F_3 = \begin{bmatrix} \dfrac{\partial F_3}{\partial x} \\ \dfrac{\partial F_3}{\partial y} \\ \dfrac{\partial F_3}{\partial z} \\ \end{bmatrix}. $$
Now, the jacobian of $\vec{F}$ is the @matrix that contains all of these gradients:
$$ \vec{J} = \nabla \vec{F} = \vec{F} = \begin{bmatrix} (\nabla F_1)^\mathsf{T} \\ (\nabla F_2)^\mathsf{T} \\ (\nabla F_3)^\mathsf{T} \\ \end{bmatrix} = \begin{bmatrix} \dfrac{\partial F_1}{\partial x} & \dfrac{\partial F_1}{\partial y} & \dfrac{\partial F_1}{\partial z}\\ \dfrac{\partial F_2}{\partial x} & \dfrac{\partial F_2}{\partial y} & \dfrac{\partial F_2}{\partial z}\\ \dfrac{\partial F_3}{\partial x} & \dfrac{\partial F_3}{\partial y} & \dfrac{\partial F_3}{\partial z}\\ \end{bmatrix}. $$
Here:
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The @rows are the @transposes of gradients of each component of $\vec{F}.$
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The @columns tell how each @coordinate direction affects all components of $\vec{F}.$
Thus the jacobian tells us how each component of $\vec{F}$ changes with every @coordinate. It can be viewed as the "gradient of a vector field."
Jacobian Related to Divergence
Given a vector field $F:$
$$ \vec{F} = \begin{bmatrix} F_1(x,y,z) \\ F_2(x,y,z) \\ F_3(x,y,z) \\ \end{bmatrix} : \mathbb{R}^3 \to \mathbb{R}^3, $$
with a jacobian
$$ \vec{J} = \nabla \vec{F} = \begin{bmatrix} \dfrac{\partial F_1}{\partial x} & \dfrac{\partial F_1}{\partial y} & \dfrac{\partial F_1}{\partial z}\\ \dfrac{\partial F_2}{\partial x} & \dfrac{\partial F_2}{\partial y} & \dfrac{\partial F_2}{\partial z}\\ \dfrac{\partial F_3}{\partial x} & \dfrac{\partial F_3}{\partial y} & \dfrac{\partial F_3}{\partial z}\\ \end{bmatrix}, $$
divergence is the @sum of the @diagonal @entries of this @matrix:
$$ \nabla \cdot \vec{F} = \dfrac{\partial F_1}{\partial x} + \dfrac{\partial F_@}{\partial y} + \dfrac{\partial F_3}{\partial z} = \tr{\vec{J}}.$$
So, algebraically, divergence is the @trace of the jacobian.
The trace measures how much the @linear-transformation represented by $\vec{J}$ stretches space along its coordinate axes - it's the infinitesimal "net expansion rate."
Geometrically, if we have a very small cube of points centered at $\vec{x_0},$ with a side length of $dx,$ then each corner of that cube corresponds to a slightly different $\vec{x} = \vec{x_0} + \vec{\Delta x}.$ If we take all the points in the box and map them through $\vec{F},$ the cube's image becomes a tiny, possibly deformed box in the $\vec{F}$-space. The way the box stretches, shears and rotates is determined by $\vec{J}.$ The change in volume of that box is determined by $\det{(\vec{J})}$ (the determinant of the Jacobian.)
For an infinitesimally small cube, the @determinant can be expanded as
$$ \det{(\vec{I} + \vec{J} \epsilon)} \approx 1 + \epsilon \tr(\vec{J}), $$
for tiny $\epsilon.$ Here, the @trace of $\vec{J},$
$$ \tr{\vec{J}} = \dfrac{\partial F_1}{\partial x} + \dfrac{\partial F_2}{\partial y} + \dfrac{\partial F_3}{\partial z} = \nabla \cdot \vec{F} $$
is exactly the divergence. Each of these terms tells us how much the points in an infinitesimally small box around $\vec{x_0}$ are stretched/compressed in each coordinate direction.
Physically, for example, if $\vec{F}$ represents a velocity field, $\dfrac{\partial F_1}{\partial x}$ says how much the $x$-component of the velocity changes as you move along $x.$ Note that if $F_1$ is constant with respect to $x,$ this term is $0,$ which means there is no stretch/compression along that axis, and if we think of the velocity being that of a fluid, the same amount of fluid will flow into any region in the $x$ direction as flows out in the $x$ direction.
Jacobian Related to Curl
Given a vector field $F:$
$$ \vec{F} = \begin{bmatrix} F_1(x,y,z) \\ F_2(x,y,z) \\ F_3(x,y,z) \\ \end{bmatrix} : \mathbb{R}^3 \to \mathbb{R}^3, $$
with a jacobian
$$ \vec{J} = \nabla \vec{F} = \begin{bmatrix} \dfrac{\partial F_1}{\partial x} & \dfrac{\partial F_1}{\partial y} & \dfrac{\partial F_1}{\partial z}\\ \dfrac{\partial F_2}{\partial x} & \dfrac{\partial F_2}{\partial y} & \dfrac{\partial F_2}{\partial z}\\ \dfrac{\partial F_3}{\partial x} & \dfrac{\partial F_3}{\partial y} & \dfrac{\partial F_3}{\partial z}\\ \end{bmatrix}, $$
We can split $\vec{J}$ into its @symmetric and @antisymmetric parts:
$$ \vec{J} = \vec{S} + \vec{A}, \quad \vec{S} = \frac{1}{2}(\vec{J} + \vec{J}^{\mathsf{T}}), \quad \vec{A} = \frac{1}{2}(\vec{J} - \vec{J}^\mathsf{T}). $$
Now,
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$\vec{S}$ encodes stretching/shearing.
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$\vec{A}$ encodes local rotation.
The components of $\vec{A}$ are:
$$ \vec{A} = \frac{1}{2} \begin{bmatrix} 0 & \dfrac{\partial F_1}{\partial y} - \dfrac{\partial F_2}{\partial x} & \dfrac{\partial F_1}{\partial z} - \dfrac{\partial F_3}{\partial x} \\ \dfrac{\partial F_2}{\partial x} - \dfrac{\partial F_1}{\partial y} & 0 & \dfrac{\partial F_2}{\partial z} - \dfrac{\partial F_3}{\partial y} \\ \dfrac{\partial F_3}{\partial x} - \dfrac{\partial F_1}{\partial z} & \dfrac{\partial F_3}{\partial y} - \dfrac{\partial F_2}{\partial z} & 0 \\ \end{bmatrix}. $$
If we let $\vec{\omega} = (\omega_x, \omega_y, \omega_z),$ we can write $\vec{A}$ as
$$ \vec{A} = \begin{bmatrix} 0 & - \omega_z & \omega_y \\ \omega_z & 0 & - \omega_x \\ - \omega_y & \omega_x & 0 \\ \end{bmatrix}. $$
Then,
$$ \vec{\omega} = \frac{1}{2} \nabla \times \vec{F}, $$
so curl is twice the @axial-vector of the @antisymmetric part of the jacobian.
Geometrically, if we linearize $\vec{F}$ near some point $\vec{x_0},$
$$ \vec{F}(\vec{x_0} + \vec{\Delta x}) \approx \vec{F(x_0)} + \vec{J(x_0)} \vec{\Delta x}. $$
The matrix $\vec{J}$ is acting on small displacements. When we apply $\vec{A}$ to $\vec{\Delta x},$ we get
$$ \vec{A} \vec{\Delta x} = \vec{\omega} \times \vec{\Delta x}. $$
Then,
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The direction of $\vec{\omega}}}$ is the axis of rotation (via the right-hand rule).
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The magnitude $||\omega||$ is the local @angular-velocity (half the curl magnitude.)
More visually, if we take a small ball of points around $\vec{x_0},$
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The symmetric part $\vec{S}$ of $\vec{J}$ turns the ball into an @ellipsoid (by stretching/shearing it).
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The antisymmetric part $\vec{A}$ spins the ball about the axis $\omega$ without changing its shape or volume.
Thus, divergence comes from the @trace of $S$ and curl comes from $A.$
Physically, if $F$ is a velocity field of a fluid, then:
$$ \nabla \times \vec{F} = 2 \omega $$
means that the fluid near that point is rotating like a tiny rigid body, with @angular-velocity $\vec{\omega.}$
So, if we dropped a tiny paddle wheel into the flow:
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The wheel's axis aligns with $\nabla \times \vec{F}.$
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Its spin rate is $\frac{1}{2} || \nabla \times \vec{F} ||.$
Jacobian Related to Directional Derivative
This is mostly covered above, so just note that the jacobian is a sort of machine for producing directional derivatives:
$$ D_\vec{v} F = \vec{J} \vec{v}, $$
where the multiplication on the right is matrix-vector multiplication, i.e. the image of $\vec{J}$ when applied to $\vec{v}.$
Jacobian Related to Laplacian
Algebraically,
$$ \nabla^2 f = \nabla \cdot (\nabla f). $$
To unpack that, we're saying that the laplacian is the divergence of the gradient of $f.$ Now, when we take the gradient of $f,$ we get a vector field that has the first partial derivatives of $f.$ Now, if we take the jacobian of that gradient, we get $\nabla (\nabla f)),$ which is the hessian of $f.$ Now, if we take the @trace of that hessian, we get the laplacian, which is the sum of the pure second @derivatives of $f.$
Geometrically, the hessian is telling us how the gradient of $f$ curves near a point, and taking the trace of it tells us the total or average @curvature at that point.
If $\nabla^2 f = 0,$ the gradient's inflow and outflow balance in every direction: the field is @harmonic and locally curvature-neutral.