Lecture 014

Vector Field: a vector function $\bar{F} : D \rightarrow \mathbb{R}^n$ where $D$ is a domain in $\mathbb{R}^n$

$\bar{F}(x_1, ..., x_n) = \langle{f_1(x_1, ..., x_n), ..., f_n(x_1, ..., x_n)}\rangle$ where $f_i$ is component function.
$\bar{F}(x, y) = \langle{P(x, y), Q(x, y)}\rangle$
$\bar{F}(x, y, z) = \langle{P(x, y, z), Q(x, y, z), R(x, y, z)}\rangle$

Unit Vector Field: $(\forall (x, y) \in D)(\|\bar{F}(x, y)\| = 1)$

example: $\bar{F}(x, y) = \langle{\frac{-y}{\sqrt{x^2 + y^2}}}, \frac{x}{\sqrt{x^2 + y^2}}\rangle$

Gradient (Conservative) Vector Field: $\nabla f(x, y, z) = \langle{f_x (x, y, z), f_y (x, y, z), f_z (x, y, z)}\rangle$

$\bar{F}$ is a gradient vector field iff $(\exists f \in \text{scalar function})(\bar{F} = \nabla f)$
Uniqueness of Potential Functions: for function $f, g$ , then $\nabla f = F = \nabla g \implies (\exists C \in \mathbb{R})(f = g + C)$
Cross-Partial Property: if $\bar{F}(x, y) = \langle{f_x = P(x, y), f_y = Q(x, y)}\rangle$ is conservative and $P, Q$ are $C^1$ continuous then $\frac{\partial P}{\partial y} = f_{xy} = f{yx} = \frac{\partial Q}{\partial x}$ .
- if $\bar{F}(x, y, z) = \langle{P(x, y, z), Q(x, y, z), R(x, y, z)}\rangle$ is conservative and $P, Q, R$ are $C^1$ continuous then $\begin{cases} \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} \implies \frac{\partial f}{\partial x \partial y} = \frac{\partial f}{\partial y \partial x}\\ \frac{\partial P}{\partial z} = \frac{\partial R}{\partial x} \implies \frac{\partial f}{\partial x \partial z} = \frac{\partial f}{\partial z \partial x}\\ \frac{\partial R}{\partial y} = \frac{\partial Q}{\partial z} \implies \frac{\partial f}{\partial z \partial y} = \frac{\partial f}{\partial y \partial z}\\ \end{cases}$ .
- If $\bar{F}(x, y)$ is conservative: then $P = f_x, Q = f_y$ , then $f_{xy} = f_{yx}$ so $\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$ (Can be used to show $\bar{F}(x, y)$ is not conservative)
- Partial Converse to Cross-Partial Property: If $\overrightarrow{F}$ is defined on a simply connected domain, then $\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} \implies \overrightarrow{F} \text{ is conservative}$
- Combine Together forms a Stronger theory: $F$ is a vector field on an open, simply connected region $D$ , then $P_y = Q_x, P_z = R_x, Q_z = R_y \iff F \text{ is conservative}$ . (a hole is domain is not simply connected)

Find potential formula: given $\bar{F}(x, y) = \langle{P, Q}\rangle = \langle{f_x, f_y}\rangle$ .

verify $\frac{\partial f_x}{\partial y} = \frac{\partial f_y}{\partial x}$ ( $f_{xy} = f_yx$ ) by verifying $\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$
find $\int f_x dx = [something_1] + g(y)$ and $\int f_y dy = [something_2] + h(x)$
find $g(y), h(x)$ to make $[something_1] + g(y) = [something_2] + h(x)$
the resulting function is potential function $f$

// QUESTION: gradient vector field is the one doesn't form loop (can be thought as potential)

Scalar Line Integral

Scalar Line Integral: given a curve $C = \overrightarrow{r}(t) = \langle{x(t), y(t)}\rangle$ is a $C^1$ -continuous function, and a scalar function $f(x, y)$

definition: line integral of $f$ over $C$ is $\int_C f dS = \lim_{n \rightarrow \inf} \sum_1^n f(P_i) \Delta S_i$ where $\Delta S_i = \int_{t_{i - 1}}^{t_i} \|\overrightarrow{r}'(t)\| dt$ is change of the arc length of the $i$ -th interval on $C$ . ( $dS = \|\overrightarrow{r}'(t)\| dt$ )
calculation: $\int f dS = \int_a^b f(r(t)) \|r'(t)\| dt$
- In the normal integration for area, we take $\|r'(t)\| = 1$ because it is the same progression of coordinates. In this case, $t$ might run very slow and thus scale the result.
arc length: $\int_C 1 dS$

Vector Line Integrals

Piecewise smooth curve: piecewise function when each function is smooth.

Vector Line Integrals: Integrate field that are aligned to the tangent of a line. (a way to measure net flow of interest in a region of vector Field)

$\begin{align*} \int_C F dS = \int_a^b F(r(t)) \cdot \overrightarrow{r'(t)} dt \end{align*}$

Circulation and Flow

Circulation: Integrate field that are aligned to the normal of a closed circle. (closed curve form a loop) Flow: how much a vector is tangential to boundary (not a loop)

Work:

$\begin{align*} &\int_C \overrightarrow{F} \cdot \overrightarrow{T} ds \tag{where $ds$ is tiny distance}\\ = &\int_C \overrightarrow{F} \cdot \frac{\overrightarrow{r}'(t)}{\|r'(t)\|} \|r'(t)\| dt\\ = &\int_C \overrightarrow{F} \cdot \overrightarrow{r}'(t) dt = \int_a^b \overrightarrow{F}(\overrightarrow{r}(t)) \cdot \overrightarrow{r}'(t) dt\\ = &\int_C \overrightarrow{F} \cdot \frac{d\overrightarrow{r}}{dt} dt\\ = &\int_C \overrightarrow{F} \cdot d \overrightarrow{r}\\ = &\int_C Pdx + Qdy\\ \end{align*}$

Alternative Notation: $\int_C Pdx + Q dy + Rdz$ for $\int_C \overrightarrow{F} d \overrightarrow{r}$ where $\overrightarrow{F} = \langle{P, Q, R}\rangle$
- This is because

$\begin{align*} &\int_C \overrightarrow{F} \cdot d \overrightarrow{r}\\ = &\int_C \overrightarrow{F} \overrightarrow{r}'(t) dt\\ = &\int_C \langle{P, Q, R}\rangle \cdot \langle{\frac{dx}{dt}, \frac{dy}{dt}, \frac{dz}{dt}}\rangle dt\\ = &\int_C Pdx + Qdy + Rdz\\ \end{align*}$

Fundamental theorem for line integrals: For any function and $C^1$ curve $C$ . $\int_C \nabla f \cdot dr = f(\overrightarrow{r}(b)) - f(\overrightarrow{r}(a))$ where $\overrightarrow{r}(t)$ for $a \leq t \leq b$ is a piece-wise parameterization of curve $C$ .

proof: $\int_a^b \nabla f \cdot \overrightarrow{r}'(t) dt = \int_a^b \frac{\partial f}{\partial x}\frac{dx}{dt} + \frac{\partial f}{\partial y} \frac{dy}{dt} + \frac{\partial f}{\partial z}\frac{dz}{ft} dt = \int_a^b \frac{df(x(t), y(t), z(t))}{dt} dt = f(\overrightarrow{r}(b)) - f(\overrightarrow{r}(a))$
Corollary: If $C$ is "closed curve" (a loop) and $\overrightarrow{F}$ is conservative, then $\int_C \overrightarrow{F} \cdot d \overrightarrow{r} = 0$ (this is because $\overrightarrow{F}$ is conservative means $\exists \nabla f = \overrightarrow{F}$ . Fix $f$ . Because there is a loop, $\int_C \nabla f \cdot d \overrightarrow{r} = \int_a^a \nabla f \cdot d \overrightarrow{r} = 0$ )

Independent path: a field is path-independent if $\int_{C1}F \cdot dr = \int_{C2}F \cdot dr$ where $C1, C2$ with the same initial and terminal points.

$\text{conservative} \implies \text{independent path}$
$\text{independent path} \land \text{domain } D \text{ of } F \text{ is open and connected} \implies \text{conservative}$

Flux

Flux: For a curve $C$ and a vector field $F$ , flux is integrated field that are aligned to the normal of the curve. It is used to calculate fluid flow accross the curve.

$\begin{align} &\int_C F \cdot N ds \tag{where $C$ is parameterized by $\overrightarrow{r}(t) = \langle{x(t), y(t)}\rangle$, $\overrightarrow{N}(t) = \frac{\overrightarrow{n}(t)}{\|\overrightarrow{n}(t)\|}$}\\ = &\int_a^b F \cdot \frac{n(t)}{\|n(t)\|} \|r'(t)\| dt \tag{where $n(t) = \langle{y'(t), -x'(t)}\rangle$}\\ = &\int_a^b F(r(t)) \cdot n(t) dt \tag{by $\|n(t)\| =\|r'(t)\| $ since $\|\langle{x', y'}\rangle\| = \|\langle{y', -x'}\rangle\|$}\\ = &\int_a^b \langle{P(r_x), Q(r_y)}\rangle \cdot \langle{y'(t), -x'(t)}\rangle dt\\ = &\int_C -Qdx + Pdy \tag{by $\langle{dx, dx}\rangle$ rotate to $\langle{dy, -dx}\rangle$}\\ \end{align}$

Source-Free: If for all closed curves $C$ on continuous vector field $\overrightarrow{F}$ on an open connected domain, $\oint_C \overrightarrow{F} \cdot \overrightarrow{N} dA = 0$ (flux = 0), then $\overrightarrow{F}$ is source-free.

Theorem: $\text{div}(\overrightarrow{F}) = 0 \iff \overrightarrow{F} \text{ is source-free}$ when $\overrightarrow{F} = \langle{P, Q}\rangle$ is a $C^1$ continuous vector field on a simply connected domain.

Source-free: on simply connected domain, the followings are true

flux of a closed curve is 0: $\oint_C F \cdot N ds = 0$
flux is path-independent
there is a stream function $g$ such that $F(a, b) \cdot \nabla g(a, b) = 0$ (similar to potential function)
$P_x + Q_y = 0$ (similar to properties of conservative vector fields)

// TODO: carefully exam when green's theorem apply (with/without holes) at the bottom of 6.4 and 6.3

Video Explaining Divergence and Curl Video Explaining Flow and Flux

Curl

Curl: for vector field $\overrightarrow{F} = \langle{P, Q, R}\rangle$ :

$\begin{align*} \text{curl}(\overrightarrow{F}) = &\nabla \times \overrightarrow{F}\\ = &\det \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z}\\ P & Q & R\\ \end{vmatrix}\\ = &\langle{R_y - Q_z, P_z - R_x, Q_x - P_y}\rangle\\ \end{align*}$

Conservative Theorem:
- if $\overrightarrow{F}$ is conservative, then $\text{curl}(\overrightarrow{F}) = 0$ (proof: $\text{conservative} \iff \begin{cases} P_z = R_x\\ P_y = Q_x\\ R_y = Q_2\\ \end{cases}$ )
- Assume $\overrightarrow{F} = \langle{P, Q, R}\rangle$ is simply connected. Then $\text{curl}(\overrightarrow{F}) = \overrightarrow{0} \iff \overrightarrow{F} \text{ is conservative}$
In 3D: each component represent rotation about one axis
In 2D: for $\overrightarrow{F} = \overrightarrow{P, Q}$ , $curl \overrightarrow{F} = \langle{0, 0, Q_x - P_y}\rangle$

Divergence

Divergence: the divergence for $\overrightarrow{F} = \langle{P, Q, R}\rangle$ is the scalar field: Divergence measure how much water flow out than flow in at given point.

$\begin{align*} \text{div}(\overrightarrow{F}) &= P_x + Q_y + R_z\\ &= \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z}\\ &= \nabla \cdot \overrightarrow{F} \end{align*}$

magnetic fields: $\text{div}(\overrightarrow{F}) = 0$
$\text{div} F = 0 \iff F \text{ is source-free}$

Green's Theorem

Simply-connected: connected piece with no hole. If you make a circle (if possible) and fill in the circle, the picture would not change.

example: a point, a line, a ball
counter example: donuts, circle

Use of notation:

$s$ : the length of a curve in $\mathbb{R}^3$
$A$ : an area in $\mathbb{R}^2$
$S$ : a surface in $\mathbb{R}^3$
$D$ : a region in $\mathbb{R}^2$
$E$ : a volume in $\mathbb{R}^3$

Green's Theorem for Calculating Circulation

Green's Theorem for non-conservative: Let $D \in \mathbb{R}^2$ be open, simply connected region with boundary $\partial D \in \mathbb{R}^2$ . The boundary $\partial D$ is piecewise $C^1$ smooth.

Let $\overrightarrow{F} = \langle{P, Q}\rangle \in \mathbb{R}^2$ (may not be conservative) Then:

$\begin{align*} &\oint_{\partial D = C} \overrightarrow{F} \cdot d \overrightarrow{r} \tag{assumed counterclockwise}\\ (= &\oint_C Pdx + Qdy)\\ = &\iint_D (Q_x - P_y) dt\\ = &\iint_D (Q_x - P_y) dA\\ = &\iint_D ((Q_x - P_y) \overrightarrow{k}) \cdot \overrightarrow{k} dA \tag{if in 2D}\\ = &\iint_D \langle{0, 0, Q_x - P_y}\rangle \cdot \overrightarrow{k} dA\\ = &\iint_D \text{curl}(F) \cdot \overrightarrow{k} dA \tag{by curl in the middle cancel neighbors out}\\ = &\iint_D \text{curl}(F) \cdot \langle{0, 0, 1}\rangle dA\\ \end{align*}$

Green's Theorem for not simply-connected: Let $D \in \mathbb{R}^2$ be region with $n \in \mathbb{N}$ many holes, then green's theorem hold.

Green's Theorem to Calculate Flux

Flux version of Green's Theorem: Let $D \in \mathbb{R}^2$ be open, simply connected region with boundary $C \in \mathbb{R}^2$ . The boundary $C$ is piecewise $C^1$ smooth. Let $\overrightarrow{F} = \langle{P, Q}\rangle \in \mathbb{R}^2$ (may not be conservative) Then $\oint_C \overrightarrow{F} \cdot \overrightarrow{N} dA = \iint_D \text{div}(\overrightarrow{F}) dA = \iint_D P_x + Q_y dA$

Proof:

$\begin{align*} &\oint_C \overrightarrow{F} \cdot \overrightarrow{N} dA\\ = &\oint_C \langle{P, Q}\rangle \cdot \langle{y', -x'}\rangle dt\\ = &\oint_C Pdy - Qdx\\ = &\oint_C -Qdx + Pdy\\ = &\iint_D (P_x - (-Q_y)) dt \tag{by ordinary Green's Theorem}\\ = &\iint_D (P_x + Q_y) dA\\ = &\iint_D \text{div}(\overrightarrow{F}) dA \tag{by curl in the middle cancel neighbors out}\\ \end{align*}$

Green's Theorem: $\oint_C F \cdot N ds = \iint_D \text{div} F dA$ (geometrically, the sum of divergence at each point in a region is the sum of divergence at border of the region which is exactly the flux)

Laplacian

Laplacian (Laplace operator): for scalar field $f$ , $\Delta$ is a second-order differential operator in Euclidean space defined as the divergence ( $\nabla \cdot$ ) of the gradient ( $\nabla f$ )

$\begin{align*} \Delta f = \nabla^2 f = &\text{div}(\nabla f)\\ =& \text{div}(\langle{f_x, f_y, f_z}\rangle)\\ =& f_{xx} + f_{yy} + f_{zz}\\ =& \sum_{i = 1}^n \frac{\partial^2 f}{\partial x_i^2} \end{align*}$

Harmonic Vector Field

Harmonic (no complex disturbance, simple flow in one direction): $f$ is harmonic iff $\Delta f = \nabla^2 f = 0$

curl-free, divergence-free, and harmonic component from Helmholtz-Hodge decomposition

Helmholtz-Hodge decomposition: decompose vector field into curl-free, divergence-free, and harmonic component.

Divergence of the Curl: Assume $\overrightarrow{F} \in \mathbb{R}^3$ is $C^2$ smooth vector field, then $\begin{cases} \text{div curl}(\overrightarrow{F}) = 0 & (\text{prove by expansion, can be used to verify existence of } \overrightarrow{F})\\ \nabla \cdot (\nabla \times \overrightarrow{F}) = 0\\ \end{cases}$

Summary

Green's Theorem: $\begin{cases} \oint_C \overrightarrow{F} \cdot d \overrightarrow{r} = \iint_D \text{curl}(\overrightarrow{F}) \cdot \langle{0, 0, 1}\rangle dA\\ \oint_C \overrightarrow{F} \cdot \overrightarrow{N} dA = \iint_D \text{div}(\overrightarrow{F}) dA\\ \end{cases}$ (in $\mathbb{R}^2$ , piecewise smooth, closed curve $C$ )
Stokes Theorem: $\oint_C \overrightarrow{F} \cdot d \overrightarrow{r} = \iint_S \text{curl} \overrightarrow{F} \cdot d \overrightarrow{S}$ (in $\mathbb{R}^3$ , piecewise smooth, closed curve $C$ , oriented $S$ )
Divergence Theorem: $\iint_S \overrightarrow{F} \cdot d \overrightarrow{S} = \iiint_E \text{div} \overrightarrow{F} dV$ (piecewise smooth oriented closed surface $S$ bounding a region $E$ in $\mathbb{R}^3$ )

Surface Integral

$r(u, v) = \langle{x(u, v), y(u, v), z(u, v)}\rangle$

Smooth Surface: $S$ parameterized by $\overrightarrow{r}(u, v)$ is a smooth surface provided that $r_u, r_v$ exists and $r_u \times r_v \neq 0$ for any value $(u, v)$

$\overrightarrow{r}(t) = \langle{1, 2}\rangle$ is not smooth because it is a point. $r(t)$ is smooth if $r'(t)$ exists and never $0$ .

Surface Area: $\overrightarrow{r}(u, v)$ for $(u, v) \in D$ is a smooth surface, then the surface area of $S$ is $\iint_D \|\overrightarrow{r}_u \times \overrightarrow{r}_v\| dA$

Scalar Surface Integral:

$\begin{align*} &\iint_S f(x, y, z) dS\\ = & \iint_D f(\overrightarrow{r}(u, v)) \|\overrightarrow{r}_u \times \overrightarrow{r}_v\| dA \tag{by $dS = \|\overrightarrow{r}_u \times \overrightarrow{r}_v\| dA$}\\ \end{align*}$

Vector Field Surface Flux Integral: where $S$ surface need to be smooth and orientable

$\begin{align*} &\iint_S \overrightarrow{F} \cdot dS\\ = &\iint_S \overrightarrow{F} \cdot \overrightarrow{N} dS\\ = &\iint_S \overrightarrow{F} \cdot \frac{\overrightarrow{r}_u \times \overrightarrow{r}_v}{\|\overrightarrow{r}_u \times \overrightarrow{r}_v\|} dS \tag{where $\overrightarrow{r}_u = \frac{\partial \overrightarrow{r}}{\partial u}, \overrightarrow{r}_v = \frac{\partial \overrightarrow{r}}{\partial v}$}\\ = &\iint_S \overrightarrow{F} \cdot (\overrightarrow{r}_u \times \overrightarrow{r}_v) dA \tag{by $dS = \|\overrightarrow{r}_u \times \overrightarrow{r}_v\| dA$}\\ \end{align*}$

Stroke's Theorem: generalization of flux version of Green's Theorem:

$S$ is a piecewise smooth oriented surface
oriented boundary $C$ is simple and closed
$\overrightarrow{F} = \langle{P, Q, R}\rangle$ is a $C^1$ vector field in $\mathbb{R}^3$ that contains $S$

$\begin{align*} &\int_C \overrightarrow{F} \cdot dr\\ = &\iint_S \text{curl}\overrightarrow{F} \cdot dS\\ = &\iint_S \text{curl}\overrightarrow{F} \cdot \overrightarrow{N} dS\\ = &\iint_S \text{curl}\overrightarrow{F} \cdot \frac{\overrightarrow{r}_u \times \overrightarrow{r}_v}{\|\overrightarrow{r}_u \times \overrightarrow{r}_v\|} \|\overrightarrow{r}_u \times \overrightarrow{r}_v\| dS\\ = &\iint_S \text{curl}\overrightarrow{F} \cdot (\overrightarrow{r}_u \times \overrightarrow{r}_v) dA\\ \end{align*}$

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