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Finished Line Integrals

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% !TeX root = ../script.tex
\documentclass[../../script.tex]{subfiles}
\begin{document}
\chapter{Integration over Submanifolds}
\vspace*{\fill}\par
\pagebreak
\subfile{sections/line_integrals.tex}
\end{document}

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% !TeX root = ../../script.tex
\documentclass[../../script.tex]{subfiles}
\begin{document}
\section{Line Integrals}
\begin{defi}
Let $I$ be an interval and $n \in \natn$. A parametrized curve (or path) in $\realn^n$ is a continuous mapping
\[
\gamma: I \longrightarrow \realn^n
\]
A parametrized curve is said to be regular if it is $C^1$ and $\gamma'(t) \ne 0 ~~\forall t \in I$.
It is said to be piecewise regular if there is a disjoint decomposition
\[
I = I_1 \cup I_2 \cup \cdots \cup I_n
\]
into partial intervals such that $\gamma$ is regular on each partial interval.
A curve is a subset of $\realn^n$ that is the image of a parametrized curve. If $\curve$ is a curve, then
\[
\gamma: I \longrightarrow \realn^n
\]
is said to be the parametrization of $\curve$, if $\gamma(I) = \curve$ and if $\gamma$ is injective on $\interior{I}$.
The curves in this chapter will always be regular.
\end{defi}
\begin{eg}
\begin{enumerate}[(i)]
\item $\alpha, \kappa > 0$:
\begin{align*}
\gamma: \realn &\longrightarrow \realn^3 \\
t &\longmapsto (\cos(\alpha t), \sin(\alpha t), \kappa t)
\end{align*}
This is the parametrization of a screw curve.
\item The unit circle
\[
\set[x^2 + y^2 = 1]{(x, y) \in \realn^2}
\]
is a curve with the parametrization
\begin{align*}
\gamma: [0, 2\pi] &\longrightarrow \realn^2 \\
t &\longmapsto (\cos t, \sin t)
\end{align*}
\item A square
\[
\set[\max\set{\abs{x_1}, \abs{x_2}} = 1]{(x, y) \in \realn^2}
\]
is a piecewise regular curve.
\end{enumerate}
\end{eg}
\begin{rem}
Let $\gamma: I \rightarrow \realn^n$ be regular, $f: \gamma(I) \rightarrow \realn$ be continuous and $a, b \in \interior{I}$.
A decomposition $Z$ is given by the grid points
\[
a = t_0 < t_1 < \cdots < t_n = b
\]
The fineness of $Z$ is given by
\[
m(Z) := \max_{t \in \set{0, 1, \cdots, n-1}} (t_{i+1} - t_i)
\]
We can represent $I$ in terms of $Z$ via
\[
I(Z) := \sum_{i=0}^{n-1} f(\gamma(t_i)) \norm{\gamma(t_{i+1}) - \gamma(t_i)}
\]
Or in integral representation
\[
I(Z) = \int_a^b \underbrace{\sum_{i=0}^{n-1} f(\gamma(t_i)) \frac{\norm{\gamma(t_{i+1}) - \gamma(t_i)}}{\norm{t_{i+1} - t_i}} \charfun_{[t_i, t_{i+1})}(t)}_{g_Z(t)} \dd{t}
\]
So let $(Z_j)$ be a sequence of decompositions with
\[
m(Z_j) \conv{j \rightarrow \infty} 0
\]
Let $t \in [a, b]$ not be a grid point of any $Z_j$. Then there exists a unique grid poiont $t_{j, i_j}$ such that $t \in [t_{j, i_j}, t_{j, i_{j+1}}]$. Then
\[
\limes{j}{\infty} t_{j, i_j} = \limes{j}{\infty} t_{j, i_{j+1}} = t
\]
And thus
\[
\limes{j}{\infty} g_{Z_j}(t) = f(\gamma(t)) \norm{\gamma'(t)}
\]
$\forall t$ that are not grid points of $Z_j$, this means tahat
\[
g_{Z_j} \conv{j \rightarrow \infty} f \norm{\gamma'}
\]
almost everywhere. The dominated convergence theorem then tells us
\[
I(Z_j) = \int_a^b g_{Z_j}(t) \dd{t} \conv{j \rightarrow \infty} \int_a^b f(\gamma(t)) \norm{\gamma'(t)} \dd{t}
\]
Special case: For $f \equiv 1$ one gets the arc length.
\end{rem}
\begin{defi}[Line Integrals, Arc Length]
Let $I$ be an interval and $\gamma: I \rightarrow \realn^n$ a parametrized curve. Define the functions
\begin{align*}
f: \gamma(I) \longrightarrow \realn && E: \gamma(I) \longrightarrow \realn^n
\end{align*}
Then
\[
\int_{\gamma} f \dd{s} := \int_I f(\gamma(t)) \norm{\gamma'(t)} \dd{t}
\]
is said to be a scalar line integral (line integral of first kind), and
\[
\int_{\gamma} \innerproduct{E}{\dd{s}} := \int_I \innerproduct{E(\gamma(t))}{\gamma'(t)} \dd{t}
\]
is said to be a vector line integral (line integral of second kind).
The function $f$ or the vector field $E$ are integrable along $\gamma$ if the according integral exists.
The integral
\[
\int_{\gamma} \dd{s}
\]
is the arc length of $\gamma$, and $\gamma$ is said to be rectifiable if this integral is finite.
If the curve $\gamma$ is closed, i.e. if $I = [a, b] ~~a, b \in \realn$ and
\[
\gamma(a) = \gamma(b)
\]
Then the above integrals are often notated as
\begin{align*}
\oint_{\gamma} \dd{s} && \oint_{\gamma} \innerproduct{E}{\dd{s}}
\end{align*}
to emphasize that the curve is closed. This changes nothing about the formulas, it is merely visual.
I will try to adhere to this style.
\end{defi}
\begin{eg}[Circumference of the unit circle]
Define
\begin{align*}
\gamma: [0, 2\pi] &\longrightarrow \realn^2 \\
t &\longrightarrow (\cos(t), \sin(t))
\end{align*}
and derive this function
\[
\gamma'(t) = (-\sin(t), \cos(t)) \implies \norm{\gamma'(t)} = 1
\]
Then the circumference is
\[
\oint_{\gamma} \dd{s} = \int_0^{2\pi} \dd{t} = 2\pi
\]
\end{eg}
\begin{rem}
\begin{enumerate}[(i)]
\item If $\gamma$ is only piecewise regular then the integrands might not be defined for all $t$.
\item Line integrals don't depend on the chosen parametrization. This means if $\curve$ is a curve and
\begin{align*}
\gamma: I \rightarrow \curve && \rho: J \rightarrow \curve
\end{align*}
are parametrizations, then
\[
\int_{\gamma} f \dd{s} = \int_{\rho} f \dd{s}
\]
We also write
\[
\int_{\curve} f \dd{s}
\]
The same holds for vector integrals.
\item Both kinds of integrals depend on the scalar product.
\item Both kinds of integrals are special cases of integrals over so called One-forms
\end{enumerate}
\end{rem}
\begin{thm}
Let $\gamma: I \rightarrow \realn^n$ be a parametrized curve, and $\vartheta: J \rightarrow I$ a diffeomorphism
(so $\vartheta \in C^1$ and $\vartheta'(t) \ne 0 ~~\forall t \in J$). Let $f: \gamma(I) \rightarrow \realn$, then
\[
\int_{\gamma} f \dd{s} = \int_{\gamma \circ \vartheta} f \dd{s}
\]
\end{thm}
\begin{proof}
We can assume $I, J$ to be open, since the endpoints of the integrals are a null set and thus don't matter.
W.l.o.g. let $\gamma$ be regular. Then
\begin{equation}
\begin{split}
\int_{\gamma \circ \vartheta} f \dd{s} &= \int_J f(\gamma \circ \vartheta)(t) \norm{(\gamma \circ \vartheta)'(t)} \dd{t} \\
&= \int_J f(\gamma(\vartheta(t))) \norm{\gamma'(\vartheta(t)) \vartheta'(t)} \dd{t} \\
&= \int_J f(\gamma(\vartheta(t))) \norm{\gamma'(\vartheta(t))} \abs{\vartheta'(t)} \dd{t} \\
&= \int_I f(\gamma(\tau)) \norm{\gamma'(\tau)} \dd{\tau} \\
&= \int_{\gamma} f \dd{s}
\end{split}
\end{equation}
\end{proof}
\begin{rem}
\begin{enumerate}[(i)]
\item One can show that for a curve $\curve$ and parametrizations
\begin{align*}
\gamma: I \rightarrow \curve && \rho: J \rightarrow \curve
\end{align*}
there exists a diffeomorphism $\vartheta: J \rightarrow I$ such that
\[
\rho = \gamma \circ \vartheta
\]
So the line integral of first degree doesn't depend on the parametrization.
\item A line integral of second degree doesn't depend on the parametrization if the parametrizations run along the curve in the same direction.
So if $\vartheta' > 0$, $\vartheta$ is said to conserve orientation. If $\vartheta' < 0$ then the integral switches sign.
\end{enumerate}
\end{rem}
\begin{eg}
Let $\gamma: I \rightarrow \realn^3$ be the trajectory of a point mass, and $F: \realn^3 \rightarrow \realn^3$ a time-independent forcefield.
The work done is then given by
\[
W := \int_{\gamma} \innerproduct{F}{\dd{s}}
\]
The fact that the parametrization can be chosen arbitrarily means that the work done in a forcefield is independent from the velocity of the point mass.
\end{eg}
\begin{rem}
\begin{enumerate}[(i)]
\item Line integrals are linear in $f$ or $E$, meaning for
\[
f, g: \gamma(I) \rightarrow \realn, ~~\lambda \in \realn
\]
we have
\[
\int_{\gamma}(g + \lambda g) \dd{s} = \int_{\gamma} f \dd{s} + \lambda \int_{\gamma} g \dd{s}
\]
\item Parametrized curves over compact intervals can be reparametrized so that $I = [0, 1]$.
\item Let
\begin{align*}
\gamma: [0, 1] \rightarrow \realn^n && \rho: [0, 1] \rightarrow \realn^n
\end{align*}
be curves with $\gamma(1) = \rho(0)$. Define
\begin{align*}
\inv{\gamma}: [0, 1] &\longrightarrow \realn^n & \gamma\rho: [0, 1] &\longrightarrow \realn^n \\
t &\longrightarrow \gamma(1 - t) & t &\longrightarrow \begin{cases}
\gamma(2t), & t \le 0.5 \\
\rho(2t + 1), & t > 0.5
\end{cases}
\end{align*}
Then we have
\begin{align*}
&\int_{\inv{\gamma}} f \dd{s} = \int_{\gamma} f \dd{s} \\
&\int_{\gamma\rho} f \dd{s} = \int_{\gamma} f \dd{s} + \int_{\rho} f \dd{s} \\
&\int_{\inv{\gamma}} \innerproduct{E}{\dd{s}} = -\int_{\gamma} \innerproduct{E}{\dd{s}} \\
&\int_{\gamma\rho} \innerproduct{E}{\dd{s}} = \int_{\gamma} \innerproduct{E}{\dd{s}} + \int_{\rho} \innerproduct{E}{\dd{s}}
\end{align*}
\end{enumerate}
\end{rem}
\begin{defi}
Let $U \subset \realn^n$ be open and $f: U \rightarrow \realn$ a $C^1$-function. Define
\[
\grad f = (\partial_1 f, \partial_2 f, \cdots, \partial_m f)
\]
The vector field $E: U \rightarrow \realn^n$ is said to be conservative if there is a function $g: U \rightarrow \realn$ such that
\[
E = \grad{g}
\]
$g$ is the potential of $E$.
\end{defi}
\begin{rem}
\begin{enumerate}[(i)]
\item In physics the sign is typically switched, so
\[
E = -\grad{g}
\]
\item The IDE
\[
p(x, y) + q(x, y)y' = 0
\]
is exact if and only if the vector field $(p, q)$ is conservative.
\item If $E$ is conservative and $C^1$, then
\[
\partial_i E_j = \partial_j E_i
\]
This condition is not sufficient in general.
\item If $g$ is a potential for $E$, then the functions
\[
g + c ~~c \in \realn
\]
are also potentials.
\item If $E$ is conservative, $g$ a potential and $\gamma: [a, b] \rightarrow \realn^n$ a curve, then
\begin{align*}
\int_{\gamma} \innerproduct{E}{\dd{s}} &= \int_a^b \innerproduct{E(\gamma(t))}{\gamma'(t)} \dd{t} \\
&= \int_a^b \left(\partial_1 g(\gamma(t)) \gamma_1'(t) + \cdots + \partial_n g(\gamma(t))\gamma_n'(t)\right) \dd{t} \\
&= \int_a^b (g \circ \gamma)'(t) \dd{t} = g(\gamma(b)) - g(\gamma(a))
\end{align*}
The vector line integral over conservative fields is independent from the chosen path (it only depends on the start and end points).
\item Let $U$ be open, path-connected and $E: U \rightarrow \realn^n$ a conservative vector field.
Choose a fixed $x_0 \in U$, and for $x \in U$ choose a parametrized curve $\gamma_x$ from $x_0$ to $x$.
Then
\[
x \longmapsto \int_{\gamma_x} \innerproduct{E}{\dd{s}}
\]
is a potential, because if $g$ is an arbitrary potential we have
\[
\int_{\gamma_x} \innerproduct{E}{\dd{s}} = g(x) - g(x_0) \quad \forall x \in U
\]
\end{enumerate}
\end{rem}
\begin{eg}
\begin{enumerate}[(i)]
\item Let
\begin{align*}
E: \realn^3 \setminus \set{0} &\longrightarrow \realn^3 \\
x &\longmapsto -\frac{x}{\norm{x}^3}
\end{align*}
This field is conservative, with the potential
\begin{align*}
\phi: \realn^3 \setminus \set{0} &\longrightarrow \realn \\
x &\longmapsto \rec{\norm{x}}
\end{align*}
\item Let
\begin{align*}
E: \realn^2 \setminus \set{0} &\longrightarrow \realn^2 \\
(x, y) &\longmapsto \left(-\frac{y}{x^2 + y^2}, \frac{x}{x^2 + y^2} \right)
\end{align*}
Then
\[
\partial_1 E_2 = \frac{1}{x^2 + y^2} - \frac{2x^2}{(x^2 + y^2)} = \frac{y^2 - x^2}{(x^2 + y^2)^2} = \partial_2 E_1
\]
We can calculate the line integral of $E$ along the unit circle
\begin{align*}
\gamma: [0, 2\pi] &\longrightarrow \realn^2 \\
t &\longmapsto (\cos t, \sin t)
\end{align*}
Then
\[
E(\gamma(t)) = (-\sin t, \cos t) = \gamma'(t)
\]
The integral is then
\[
\int_{\gamma} \innerproduct{E}{\dd{s}} = \int_0^{2\pi} \norm{(-\sin t, \cos t)}^2 \dd{t} = 2\pi \ne 0
\]
\item In the chapter about differential equations we looked at an exact equation in \Cref{eg:817}:
\[
(2x + y^2) + (2xy)y' = 0
\]
We can now use curve integrals to calculate the potential function more easily. For that let $x_0 = (0, 0)$.
Then for $(\xi, \eta)$ we can define a curve connecting $x_0$ and $(\xi, \eta)$ for $t \in [0, 1]$:
\[
t \longmapsto (\xi t, \eta t)
\]
Consider the vector field
\[
E(x, y) = (2x + y^2, 2xy)
\]
Then
\begin{align*}
(\xi, \eta) \longmapsto \int_{\gamma} \innerproduct{E}{\dd{s}} &= \int_0^1 \innerproduct{E(\xi t, \eta t)}{(\xi, \eta)} \dd{t} \\
&= \int_0^1 (2\xi^2 t + \eta^2\xi t^2 + 2\xi\eta^2 t^2) \dd{t} \\
&= \xi^2 + \eta^2\xi
\end{align*}
\end{enumerate}
\end{eg}
\begin{thm}
Let $U \subset \realn^n$ be an open subset. A continuous vector field $E: U \rightarrow \realn^n$ is conservative if and only if
for every closed curve $\gamma: [0, 1] \rightarrow U$ the following holds
\[
\oint_{\gamma} \innerproduct{E}{\dd{s}} = 0
\]
\end{thm}
\begin{proof}
Line integrals over $E$ are path independent. Let $\gamma, \rho: [0, 1] \rightarrow U$ be paths with
\begin{align}
\gamma(0) = \rho(0) && \gamma(1) = \rho(1)
\end{align}
Then $\gamma \inv{\rho}$ is closed, so
\begin{equation}
0 = \int_{\gamma\inv{\rho}} \innerproduct{E}{\dd{s}} = \int_{\gamma} \innerproduct{E}{\dd{s}} - \int_{\rho} \innerproduct{E}{\dd{s}}
\end{equation}
Assume that $U$ is path continuous. Choose a fixed $x_0 \in U$ and let $g: U \rightarrow \realn$. Then
\begin{equation}
g(x) = \int_{x_0}^x \innerproduct{E}{\dd{s}}
\end{equation}
Performing a directional derivation in direction $h \in \realn^n$ yields
\begin{equation}
\begin{split}
g(x + ah) - g(x) &= \int_{x_0}^{x + ah} \innerproduct{E}{\dd{s}} - \int_{x_0}^x \innerproduct{E}{\dd{s}} \\
&= \int_{x}^{x + ah} \innerproduct{E}{\dd{s}} \\
&= \int_0^a \innerproduct{E(x + th)}{h} \dd{t}
\end{split}
\end{equation}
Here we have chosen a linear path of integration between $x_0$ and $x$, and between $x$ and $x + ah$. In other words, we're integrating along
\begin{equation}
t \longmapsto x + th
\end{equation}
Using the intermediate value theorem, we can find that $\exists \xi_a \in (0, a)$ such that
\begin{equation}
\int_0^a \innerproduct{E(x + th)}{h} \dd{t} = \innerproduct{E(x + \xi_a h)}{h} \cdot a
\end{equation}
Then we have
\begin{equation}
\partial_h g(x) = \limes{a}{0} \frac{g(x + ah) - g(x)}{a} = \limes{a}{0} \innerproduct{E(x + \xi_a h)}{h} = \innerproduct{E(x)}{h}
\end{equation}
So if $h$ is a standard basis $e_i$, then
\begin{equation}
\partial_i g(x) = E_i(x)
\end{equation}
Thus the partial derivative of $g$ is continuous, and therefore $g$ is continuously differentiable, and thus a potential.
\end{proof}
\end{document}

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The above condition is merely necessary! However, for "nice" $D$ it can be considered sufficient.
\end{rem}
\begin{eg}
\begin{eg}\label{eg:817}
Consider
\begin{align*}
\underbrace{(2x + y^2)}_p + \underbrace{2xyy'}_q = 0 && y(1) = 1

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\documentclass[11pt]{report}
\usepackage{amsmath, amssymb, amstext, physics}
\usepackage{amsmath, amssymb, amstext}
\usepackage{amsthm, stackrel, xifthen, mathtools, graphicx}
\usepackage[makeroom]{cancel}
\usepackage{hyperref, cleveref, bbm}
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\usepackage{kbordermatrix}
\usepackage{fancyhdr}
\usepackage{pdfpages}
\usepackage[arrowdel]{physics}
\usetikzlibrary{calc,trees,positioning,arrows,fit,shapes,angles,patterns}
\DeclareDocumentCommand\vnabla{}{\vectorarrow{\nabla}}
\graphicspath{assets}
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\newcommand{\setfam}{\mathcal{A}}
\newcommand{\setfamb}{\mathcal{B}}
\newcommand{\curve}{\mathcal{C}}
\newcommand{\measureable}{(\Omega, \setfam)}
\newcommand{\measure}{(\Omega, \setfam, \mu)}
\newcommand{\intervals}{\mathcal{I}}
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{\,\middle|\, #1}%
\right\}%
}
\renewcommand{\innerproduct}[2]{\langle#1,#2\rangle}
\newcommand{\equalexpl}[1]{%
\underset{\substack{\big\uparrow\\\mathrlap{\text{\hspace{-1.5em}#1}}}}{=}}
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\subfile{chapters/multivar_calc.tex}
\subfile{chapters/measures_integrals.tex}
\subfile{chapters/ode.tex}
\subfile{chapters/int_submanifold.tex}
\end{document}