Vector Calculus

Let $U\subseteq\mathbb{R}^n$ be open. A scalar field on $U$ is a function $f:U\rightarrow\mathbb{R}$ and a vector field on $U$ is a function $\underline{F}=\left(F_1,\ldots,F_n\right):U\rightarrow\mathbb{R}^n$ that assigns a vector to each point. A vector field is $C^k$ if each of its components is. For $\underline{x},\underline{y}\in\mathbb{R}^n$ the dot product

$$\underline{x}\cdot\underline{y}:=\sum_{i=1}^{n}x_i\,y_i=\underline{x}^\intercal\underline{y}$$

is the standard inner product on $\mathbb{R}^n,$ so that $\lVert\underline{x}\rVert=\sqrt{\underline{x}\cdot\underline{x}}$ is the Euclidean distance from $\underline{x}$ to $\underline{0}$ and two vectors are orthogonal if and only if their dot product is zero. The directional derivative of a differentiable scalar field is $D_{\underline{v}}f(\underline{x})=\nabla f(\underline{x})\cdot\underline{v}.$

A curve in $\mathbb{R}^n$ is a function $\underline{\gamma}=\left(\gamma_1,\ldots,\gamma_n\right):\left[a,b\right]\rightarrow\mathbb{R}^n$ with continuous components, whose image is traced from the initial point $\underline{\gamma}(a)$ to the terminal point $\underline{\gamma}(b).$ The curve is closed if $\underline{\gamma}(a)=\underline{\gamma}(b)$ and simple if it is injective on $\left[a,b\right)$ and on $\left(a,b\right].$ A $C^1$ curve has velocity $\underline{\gamma}'(t)=\left(\gamma_1'(t),\ldots,\gamma_n'(t)\right),$ speed $\lVert\underline{\gamma}'(t)\rVert$ and, where the speed is non-zero, unit tangent vector $\underline{T}(t):=\frac{\underline{\gamma}'(t)}{\lVert\underline{\gamma}'(t)\rVert}.$ The curve is regular if its speed is never zero and piecewise $C^1$ if $\left[a,b\right]$ is the union of finitely many intervals on each of which it is $C^1.$ The arc length of a piecewise $C^1$ curve is the integral of its speed.

$$L\left(\underline{\gamma}\right):=\int_a^b\lVert\underline{\gamma}'(t)\rVert\,dt$$

Let $\varphi:\left[\alpha,\beta\right]\rightarrow\left[a,b\right]$ be a $C^1$ bijection with $\varphi'\neq 0.$ The curve $\underline{\gamma}\circ\varphi$ is a reparametrisation of $\underline{\gamma},$ which preserves orientation if $\varphi'>0$ and reverses orientation if $\varphi'<0;$ the reversed curve $\underline{\gamma}^-(t):=\underline{\gamma}(a+b-t)$ traces the same image in the opposite direction. Arc length is unchanged by reparametrisation.

Let $\underline{a},\underline{b}\in\mathbb{R}^3.$ The cross product of $\underline{a}$ and $\underline{b}$ is the vector

$$\underline{a}\times\underline{b}:=\left(a_2 b_3-a_3 b_2,\,a_3 b_1-a_1 b_3,\,a_1 b_2-a_2 b_1\right)=\det\begin{bmatrix}\underline{e}_1 & \underline{e}_2 & \underline{e}_3 \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3\end{bmatrix}$$

where the determinant is expanded along its first row of standard basis vectors. The cross product is bilinear and anticommutative, $\underline{b}\times\underline{a}=-\underline{a}\times\underline{b},$ so $\underline{a}\times\underline{a}=\underline{0},$ and it is orthogonal to both of its arguments. If $\theta$ is the angle between non-zero $\underline{a}$ and $\underline{b}$ then

$$\lVert\underline{a}\times\underline{b}\rVert=\lVert\underline{a}\rVert\lVert\underline{b}\rVert\sin\theta,$$

the area of the parallelogram spanned by $\underline{a}$ and $\underline{b},$ so $\underline{a}\times\underline{b}=\underline{0}$ if and only if $\underline{a}$ and $\underline{b}$ are linearly dependent. The standard basis satisfies $\underline{e}_1\times\underline{e}_2=\underline{e}_3,$ $\underline{e}_2\times\underline{e}_3=\underline{e}_1$ and $\underline{e}_3\times\underline{e}_1=\underline{e}_2,$ which fixes the direction of $\underline{a}\times\underline{b}$ by the right-hand rule. The scalar triple product of $\underline{a},\underline{b},\underline{c}\in\mathbb{R}^3$ is the determinant of the matrix with these rows, and its absolute value is the volume of the parallelepiped they span.

$$\underline{a}\cdot\left(\underline{b}\times\underline{c}\right)=\det\begin{bmatrix}a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3\end{bmatrix}=\underline{b}\cdot\left(\underline{c}\times\underline{a}\right)=\underline{c}\cdot\left(\underline{a}\times\underline{b}\right)$$

The vector triple product satisfies $\underline{a}\times\left(\underline{b}\times\underline{c}\right)=\left(\underline{a}\cdot\underline{c}\right)\underline{b}-\left(\underline{a}\cdot\underline{b}\right)\underline{c}.$

Let $f:U\rightarrow\mathbb{R}$ be a $C^1$ scalar field and $\underline{F}:U\rightarrow\mathbb{R}^n$ be a $C^1$ vector field. The del operator $\nabla:=\left(\frac{\partial}{\partial x_1},\ldots,\frac{\partial}{\partial x_n}\right)$ is the vector of partial derivative operators, so that the gradient $\nabla f$ is the vector field of partial derivatives of $f.$ The gradient at $\underline{x}$ points in the direction in which $f$ increases fastest, since $D_{\underline{v}}f(\underline{x})=\nabla f(\underline{x})\cdot\underline{v}$ is largest over unit vectors $\underline{v}$ when $\underline{v}$ is parallel to $\nabla f(\underline{x}),$ and its length $\lVert\nabla f(\underline{x})\rVert$ is that largest rate of increase. For $c\in\mathbb{R}$ the level set of $f$ at $c$ is $\left\{\underline{x}\in U:f(\underline{x})=c\right\},$ and if $\underline{\gamma}$ is a $C^1$ curve in a level set then the chain rule gives $\nabla f\left(\underline{\gamma}(t)\right)\cdot\underline{\gamma}'(t)=0,$ so the gradient is orthogonal to level sets.

The divergence of $\underline{F}$ is the scalar field given by the dot product of $\nabla$ with $\underline{F}.$

$$\nabla\cdot\underline{F}:=\sum_{i=1}^{n}\frac{\partial F_i}{\partial x_i}=\mathrm{tr}\,J_{\underline{F}}$$

The divergence is also written $\mathrm{div}\,\underline{F},$ and $\underline{F}$ is solenoidal if $\nabla\cdot\underline{F}=0$ on $U.$ When $n=3$ the curl of $\underline{F}$ is the vector field given by the cross product of $\nabla$ with $\underline{F}.$

$$\nabla\times\underline{F}:=\left(\frac{\partial F_3}{\partial x_2}-\frac{\partial F_2}{\partial x_3},\,\frac{\partial F_1}{\partial x_3}-\frac{\partial F_3}{\partial x_1},\,\frac{\partial F_2}{\partial x_1}-\frac{\partial F_1}{\partial x_2}\right)$$

The curl is also written $\mathrm{curl}\,\underline{F},$ and $\underline{F}$ is irrotational if $\nabla\times\underline{F}=\underline{0}$ on $U.$ When $n=2$ the scalar curl of $\underline{F}=\left(F_1,F_2\right)$ is $\frac{\partial F_2}{\partial x_1}-\frac{\partial F_1}{\partial x_2},$ the third component of the curl of $\left(F_1,F_2,0\right).$ The divergence and curl of $\underline{F}$ are the trace and the antisymmetric part of its Jacobian matrix: $J_{\underline{F}}-J_{\underline{F}}^\intercal$ has the components of $\nabla\times\underline{F}$ as its off-diagonal entries. A vector field is radial if $\underline{F}(\underline{x})=g\left(\lVert\underline{x}\rVert\right)\underline{x}$ for a scalar function $g,$ and a $C^1$ radial field is irrotational.

Let $f:U\rightarrow\mathbb{R}$ be $C^2.$ The Laplacian of $f$ is the divergence of its gradient, the sum of its unmixed second-order partial derivatives.

$$\Delta f:=\nabla\cdot\nabla f=\sum_{i=1}^{n}\frac{\partial^2 f}{\partial x_i^2}=\mathrm{tr}\,H_f$$

The Laplacian is also written $\nabla^2 f,$ and $f$ is harmonic on $U$ if $\Delta f=0$ on $U.$ The vector Laplacian $\Delta\underline{F}$ of a $C^2$ vector field is the vector field whose components are the Laplacians of the components of $\underline{F}.$ By Schwarz’s theorem, for $C^2$ scalar fields $f,g$ and $C^2$ vector fields $\underline{F},\underline{G}$ on $U\subseteq\mathbb{R}^3$ the following vector identities hold.

$$\nabla\left(f\,g\right)=f\,\nabla g+g\,\nabla f$$ $$\nabla\cdot\left(f\,\underline{F}\right)=\nabla f\cdot\underline{F}+f\,\nabla\cdot\underline{F},\qquad\nabla\times\left(f\,\underline{F}\right)=\nabla f\times\underline{F}+f\,\nabla\times\underline{F}$$ $$\nabla\cdot\left(\underline{F}\times\underline{G}\right)=\underline{G}\cdot\left(\nabla\times\underline{F}\right)-\underline{F}\cdot\left(\nabla\times\underline{G}\right)$$ $$\nabla\times\left(\nabla f\right)=\underline{0},\qquad\nabla\cdot\left(\nabla\times\underline{F}\right)=0$$ $$\nabla\times\left(\nabla\times\underline{F}\right)=\nabla\left(\nabla\cdot\underline{F}\right)-\Delta\underline{F}$$

The first identity on the fourth line states that gradient fields are irrotational and the second that curl fields are solenoidal. The identities on the first two lines hold in $\mathbb{R}^n$ with the curl omitted.

Let $\underline{\gamma}:\left[a,b\right]\rightarrow U$ be a piecewise $C^1$ curve, $f:U\rightarrow\mathbb{R}$ be a continuous scalar field and $\underline{F}:U\rightarrow\mathbb{R}^n$ be a continuous vector field. The line integral of $f$ along $\underline{\gamma}$ with respect to arc length and the line integral of $\underline{F}$ along $\underline{\gamma}$ are

$$\int_{\underline{\gamma}}f\,ds:=\int_a^b f\left(\underline{\gamma}(t)\right)\lVert\underline{\gamma}'(t)\rVert\,dt,\qquad\int_{\underline{\gamma}}\underline{F}\cdot d\underline{r}:=\int_a^b\underline{F}\left(\underline{\gamma}(t)\right)\cdot\underline{\gamma}'(t)\,dt.$$

The second integrates the component of $\underline{F}$ along the unit tangent vector, $\int_{\underline{\gamma}}\underline{F}\cdot d\underline{r}=\int_{\underline{\gamma}}\underline{F}\cdot\underline{T}\,ds,$ and is also written $\int_{\underline{\gamma}}F_1\,dx_1+\cdots+F_n\,dx_n.$ Both line integrals are unchanged by orientation-preserving reparametrisation, and reversing the curve leaves the first unchanged and changes the sign of the second. The line integral of $\underline{F}$ around a closed curve is the circulation of $\underline{F}$ and is written $\oint_{\underline{\gamma}}\underline{F}\cdot d\underline{r}.$

A vector field $\underline{F}$ on $U$ is conservative if $\underline{F}=\nabla\varphi$ for a $C^1$ scalar field $\varphi:U\rightarrow\mathbb{R},$ which is a potential of $\underline{F}.$ The gradient theorem states that the line integral of a gradient field depends only on the endpoints of the curve.

$$\int_{\underline{\gamma}}\nabla\varphi\cdot d\underline{r}=\varphi\left(\underline{\gamma}(b)\right)-\varphi\left(\underline{\gamma}(a)\right)$$

Suppose that any two points of $U$ can be joined by a piecewise $C^1$ curve in $U.$ Then two potentials of $\underline{F}$ differ by a constant and the following are equivalent for a continuous vector field $\underline{F}$ on $U:$ $\underline{F}$ is conservative; the line integral of $\underline{F}$ is path independent, taking the same value along any two curves in $U$ with the same initial and terminal points; the circulation of $\underline{F}$ around every closed piecewise $C^1$ curve in $U$ is zero. Since gradient fields are irrotational, a $C^1$ conservative field on $U\subseteq\mathbb{R}^3$ is irrotational, and a $C^1$ conservative field on $U\subseteq\mathbb{R}^2$ has zero scalar curl. The converse fails in general: $\underline{F}(x,y)=\frac{\left(-y,x\right)}{x^2+y^2}$ has zero scalar curl on $\mathbb{R}^2\setminus\left\{\underline{0}\right\}$ but circulation $2\pi$ around the unit circle. The set $U$ is simply connected if every closed curve in $U$ can be continuously shrunk within $U$ to a point, which holds for $\mathbb{R}^n,$ for open rectangles and for $\mathbb{R}^3$ with a point removed, but not for $\mathbb{R}^2$ with a point removed. On a simply connected open set every $C^1$ irrotational field is conservative.

Let $D\subset\mathbb{R}^2$ be a closed bounded region of the kind over which multiple integrals are taken and $\underline{r}:D\rightarrow\mathbb{R}^3$ be $C^1.$ The image $S=\underline{r}(D)$ is a parametrised surface with parameters $\left(u,v\right)\in D,$ and the partial derivatives $\underline{r}_u:=\frac{\partial\underline{r}}{\partial u}$ and $\underline{r}_v:=\frac{\partial\underline{r}}{\partial v}$ are tangent to $S.$ The normal vector $\underline{r}_u\times\underline{r}_v$ is orthogonal to $S,$ the surface is regular if the normal vector is never zero, and its unit normal is $\underline{n}:=\frac{\underline{r}_u\times\underline{r}_v}{\lVert\underline{r}_u\times\underline{r}_v\rVert}.$ Since $\lVert\underline{r}_u\times\underline{r}_v\rVert$ is the area of the parallelogram spanned by the tangent vectors, the surface area of $S$ and the surface integral of a continuous scalar field $f$ over $S$ are

$$A\left(S\right):=\int_D\lVert\underline{r}_u\times\underline{r}_v\rVert\,d(u,v),\qquad\int_S f\,dS:=\int_D f\left(\underline{r}(u,v)\right)\lVert\underline{r}_u\times\underline{r}_v\rVert\,d(u,v).$$

For example, the graph $\underline{r}(u,v)=\left(u,v,g(u,v)\right)$ of a $C^1$ function $g$ has $\lVert\underline{r}_u\times\underline{r}_v\rVert=\sqrt{1+g_u^2+g_v^2},$ and the sphere of radius $\rho$ parametrised by $\underline{r}(\theta,\phi)=\rho\left(\sin\phi\cos\theta,\sin\phi\sin\theta,\cos\phi\right)$ on $\left[0,2\pi\right]\times\left[0,\pi\right]$ has $\lVert\underline{r}_\theta\times\underline{r}_\phi\rVert=\rho^2\sin\phi$ and surface area $4\pi\rho^2.$

A surface is orientable if it admits a continuous choice of unit normal, and a choice of one of the two such fields $\pm\underline{n}$ is an orientation of the surface; the Möbius strip is not orientable. The flux of a continuous vector field $\underline{F}$ through an oriented surface $S$ is the surface integral of its normal component.

$$\int_S\underline{F}\cdot\underline{n}\,dS=\int_D\underline{F}\left(\underline{r}(u,v)\right)\cdot\left(\underline{r}_u\times\underline{r}_v\right)d(u,v)$$

Surface area and surface integrals of scalar fields are unchanged by reparametrisation of $S,$ and the flux changes sign when the orientation is reversed. A surface is closed if it encloses a bounded region of $\mathbb{R}^3,$ such as a sphere, and its outward normal is the unit normal that points away from that region. Flux through a closed surface is written $\oint_S\underline{F}\cdot\underline{n}\,dS.$

The Jordan curve theorem states that a simple closed curve in $\mathbb{R}^2$ separates the plane into exactly two regions, one bounded and one unbounded, of which it is the common boundary. Let $\underline{\gamma}$ be a piecewise $C^1$ simple closed curve traversed anticlockwise, so that the bounded region $D$ it encloses lies to its left, and write $\partial D$ for the curve $\underline{\gamma}.$ Green’s theorem states that for a $C^1$ vector field $\underline{F}=\left(F_1,F_2\right)$ on an open set containing $D$ and $\underline{\gamma}$

$$\oint_{\partial D}F_1\,dx+F_2\,dy=\int_D\left(\frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y}\right)d(x,y).$$

The circulation around the boundary equals the integral of the scalar curl over the region. Taking $\underline{F}=\frac{1}{2}\left(-y,x\right)$ gives the area of $D$ as $\frac{1}{2}\oint_{\partial D}x\,dy-y\,dx.$

Let $S$ be a regular oriented parametrised surface in $\mathbb{R}^3$ with unit normal $\underline{n},$ whose boundary curve $\partial S$ is the image of the boundary of $D$ under $\underline{r},$ a piecewise $C^1$ closed curve oriented so that $S$ lies to the left when walking along $\partial S$ with head in the direction of $\underline{n}.$ Stokes’ theorem states that for a $C^1$ vector field $\underline{F}$ on an open set containing $S$

$$\oint_{\partial S}\underline{F}\cdot d\underline{r}=\int_S\left(\nabla\times\underline{F}\right)\cdot\underline{n}\,dS.$$

The circulation around the boundary equals the flux of the curl through the surface. Green’s theorem is the case of a flat surface in the plane with $\underline{n}=\underline{e}_3,$ and the flux of a curl through a closed surface is zero.

Let $V\subset\mathbb{R}^3$ be a bounded region enclosed by a closed, piecewise regular surface $\partial V$ with outward normal $\underline{n}.$ The divergence theorem states that for a $C^1$ vector field $\underline{F}$ on an open set containing $V$ and $\partial V$

$$\oint_{\partial V}\underline{F}\cdot\underline{n}\,dS=\int_V\nabla\cdot\underline{F}\,d\underline{x}.$$

The flux out of the region equals the integral of the divergence over the region. The theorem holds in $\mathbb{R}^n$ with the surface integral replaced by the integral over the boundary with respect to $\left(n-1\right)$-dimensional area; in $\mathbb{R}^2$ it states that $\oint_{\partial D}\underline{F}\cdot\underline{n}\,ds=\int_D\nabla\cdot\underline{F}\,d(x,y),$ which is Green’s theorem applied to $\left(-F_2,F_1\right).$ Applied to $f\,\nabla g$ for $C^2$ scalar fields $f,g,$ it gives Green’s identities.

$$\int_V\left(f\,\Delta g+\nabla f\cdot\nabla g\right)d\underline{x}=\oint_{\partial V}f\,\nabla g\cdot\underline{n}\,dS$$ $$\int_V\left(f\,\Delta g-g\,\Delta f\right)d\underline{x}=\oint_{\partial V}\left(f\,\nabla g-g\,\nabla f\right)\cdot\underline{n}\,dS$$

Let $B_\rho=\left\{\underline{y}\in\mathbb{R}^3:\lVert\underline{y}-\underline{x}\rVert\leq\rho\right\}$ be the ball of radius $\rho$ centred at $\underline{x},$ with volume $\lvert B_\rho\rvert=\frac{4}{3}\pi\rho^3$ and boundary sphere $\partial B_\rho.$ By the divergence and Stokes’ theorems with the mean value theorem for integrals, the divergence of a $C^1$ field is its outward flux per unit volume and each component of its curl is its circulation per unit area, where $\underline{\gamma}_\rho$ is the circle of radius $\rho$ centred at $\underline{x}$ in the plane orthogonal to the unit vector $\underline{v}$ and traversed anticlockwise about $\underline{v}.$

$$\nabla\cdot\underline{F}(\underline{x})=\lim_{\rho\rightarrow 0}\frac{1}{\lvert B_\rho\rvert}\oint_{\partial B_\rho}\underline{F}\cdot\underline{n}\,dS,\qquad\left(\nabla\times\underline{F}(\underline{x})\right)\cdot\underline{v}=\lim_{\rho\rightarrow 0}\frac{1}{\pi\rho^2}\oint_{\underline{\gamma}_\rho}\underline{F}\cdot d\underline{r}$$

These limits define the divergence and curl independently of coordinates, so both are unchanged by rotations of the axes.