commit d27e2526a41b97ed9625081e935e1d99b776bf6c
Author: miksa234 <milutin@popovic.xyz>
Date: Wed, 10 Nov 2021 11:31:58 +0100
initial
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diff --git a/appl_ana/sesh1/main.pdf b/appl_ana/sesh1/main.pdf
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diff --git a/appl_ana/sesh1/main.tex b/appl_ana/sesh1/main.tex
@@ -0,0 +1,342 @@
+\documentclass[a4paper]{article}
+
+
+\usepackage[T1]{fontenc}
+\usepackage[utf8]{inputenc}
+\usepackage{mlmodern}
+
+%\usepackage{ngerman} % Sprachanpassung Deutsch
+
+\usepackage{graphicx}
+\usepackage{geometry}
+\geometry{a4paper, top=15mm}
+
+\usepackage{subcaption}
+\usepackage[shortlabels]{enumitem}
+\usepackage{amssymb}
+\usepackage{amsthm}
+\usepackage{mathtools}
+\usepackage{braket}
+\usepackage{bbm}
+\usepackage{graphicx}
+\usepackage{float}
+\usepackage{yhmath}
+\usepackage{tikz}
+\usetikzlibrary{patterns,decorations.pathmorphing,positioning}
+\usetikzlibrary{calc,decorations.markings}
+
+%\usepackage[backend=biber, sorting=none]{biblatex}
+%\addbibresource{uni.bib}
+
+\usepackage[framemethod=TikZ]{mdframed}
+
+\tikzstyle{titlered} =
+ [draw=black, thick, fill=white,%
+ text=black, rectangle,
+ right, minimum height=.7cm]
+
+
+\usepackage[colorlinks=true,naturalnames=true,plainpages=false,pdfpagelabels=true]{hyperref}
+\usepackage[parfill]{parskip}
+\usepackage{lipsum}
+
+
+\usepackage{tcolorbox}
+\tcbuselibrary{skins,breakable}
+
+\pagestyle{myheadings}
+
+\markright{Popovic\hfill Applied Analysis\hfill}
+
+
+\title{University of Vienna\\ Faculty of Mathematics\\
+\vspace{1cm}Applied Analysis Problems
+}
+\author{Milutin Popovic}
+
+\begin{document}
+\maketitle
+\tableofcontents
+
+\section{Sheet 1}
+
+\subsection{Fall from high}
+We consider a free fall ($\dot{x}(t=0)=0$) of an object with mass $20\
+\text{kg}$ from a height
+$x(0) = h = 20\; \text{km}$, such that the gravitational force depends on the hight $x(t)$ in
+the following way
+\begin{align}\label{eq: free fall}
+ \ddot{x} = -g\frac{R^2}{(x(t) + R)^2},
+\end{align}
+where $R$ is the radius of the earth $R \approx 6000\; \text{km}$ and $g
+\approx 9.91\ \frac{m}{s^2}$ is the gravitational acceleration on the surface
+of the earth. For this problem there are two possible non-dimensionalisations,
+but first let us rewrite the variables in terms of non-dimensional variables
+and some dimensional constants, a priori let
+\begin{align}
+ t &= t_c \tau \;\;\; \text{and}\\
+ x &= x_c \xi.
+\end{align}
+With the above ansatz we get the following second derivative in
+time
+\begin{align}
+ \frac{d^2}{dt^2} &= \frac{1}{t_c^2}\frac{d^2}{d\tau^2} \\
+ \Rightarrow \frac{d^2x}{dt^2} &= \frac{x_c}{t_c^2}
+ \frac{d^2\xi}{d\tau^2},
+\end{align}
+and thus the initial conditions can be rewritten as
+\begin{align}
+ \xi(0) = \frac{h}{x_c},\\
+ \dot{\xi} = 0.
+\end{align}
+Now we can rewrite the equation of the free fall in \ref{eq: free fall} in
+terms of $\xi(\tau)$ as
+\begin{align}
+ \frac{x_c}{gt_c^2} \ddot{\xi} = -\frac{1}{(\frac{x_c}{R}\xi +1)^2}.
+\end{align}
+Thereby we have three dimensional constants constants $\Pi_1, \Pi_2, \Pi_3$,
+as follows
+\begin{align}
+ \Pi_1 = \frac{x_c}{R}, \;\;\;\; \Pi_2 = \frac{h}{x_c}, \;\;\;\;
+ \Pi_3 = \frac{x_c}{gt_c^2}.
+\end{align}
+
+The first scaling is done by reducing $\Pi_1$ and $\Pi_3$ to 1, by setting
+\begin{align}
+ x_c = R, \;\;\;\; t_c = \sqrt{\frac{R}{g}},
+\end{align}
+refolmulating the initial problem in equation \ref{eq: free fall} to
+\begin{align}
+ &\ddot{\xi} = -\frac{1}{(\xi + 1)^2},\;\;\;\;
+ \text{with} \nonumber\\
+ &\xi(0) = \frac{h}{R}, \;\;\;\; \dot{\xi}(0) = 0.
+\end{align}
+Reducing the problem, meaining letting $R \rightarrow \infty$ makes the first
+initial condition $\xi(0) \rightarrow 0$. We can conclude that this scaling
+is bad since it changes the initial condition in the reduced problem.
+
+The second scaling option reduces $\Pi_2$ and $\Pi_3$ to 1, by setting
+\begin{align}
+ x_c = h, \;\;\;\; t_c = \sqrt{\frac{h}{g}},
+\end{align}
+refolmulating the initial problem in equation \ref{eq: free fall} to
+\begin{align}
+ &\ddot{\xi} = -\frac{1}{(\frac{h}{R}\xi + 1)^2},\;\;\;\;
+ \text{with} \nonumber\\
+ &\xi(0) = 1, \;\;\;\; \dot{\xi}(0) = 0.
+\end{align}
+By letting $R \rightarrow \infty$ we get the following reduced problem
+\begin{align}\label{eq: free fall reduced}
+ \ddot{\xi} = -1.
+\end{align}
+Integrating and solving for $\xi(\tau = \frac{T}{t_c}) = 0$ for when the
+object hits the ground we get a familliar solution
+\begin{align}
+ T = \sqrt{\frac{2h}{g}}
+\end{align}
+Now in the reduced problem the time untill the object hits the ground is
+shorter since the acceleration is constant, but in the original one the
+acceleration increases as the object comes closer to earth.
+Additionally we can calculate the velocity at impact we need to integrate the
+reduced problem \ref{eq: free fall reduced} once and put in the initial
+condition
+\begin{align}
+ \dot{\xi}(\tau = \frac{T}{t_c}) &= -\tau = -\sqrt{2} \\
+ \text{and} \;\;\; \dot{x} &= \frac{x_c}{t_c}\dot{\xi} =
+ \sqrt{gh}\; \dot{\xi}\\
+ \Rightarrow \dot{x}(T) &= -\sqrt{2gh}
+\end{align}
+The vectical throw allows for different scaling because the initial
+conditions are different, and thus the solution too $x(0) = 0$ and
+$\dot{x}(0) = v$.
+
+\subsection{Scaling The Van der Pol equation}
+The Van der Pol equation is a perturbation of the oscillation equation
+\begin{align}\label{eq: vanderpol}
+ LC\frac{d^2I}{dt^2} + (-g_1C +3g_3CI^2)\frac{dI}{dt} = -I
+\end{align}
+with initial conditions
+\begin{align}\label{eq: van initial}
+ I(0) = I_0,\;\;\;\; \dot{I}(0) = 0.
+\end{align}
+where $I(t)$ is the current at a time $t$, $C$ is the capacity, $L$ is the
+inductivity and $g_1, g_3$ are some parameters. The units of all the
+parameters are
+\begin{align}
+ [LC] &= s^2\\
+ [g_1C] &= s\\
+ [g_3C] &= sA^{-2}
+\end{align}
+The oscillation equation is
+\begin{align}
+ CL\ddot{I} + I = 0.
+\end{align}
+Solvable by the exponential ansatz of $I = Ae^{\lambda t}$, where $\lambda=
+\pm i \sqrt{\frac{1}{LC}}$, thereby
+\begin{align}
+ I(t) = A_1 e^{i\sqrt{\frac{1}{LC}}t} + A_2 e^{-i\sqrt{\frac{1}{LC}}t}.
+\end{align}
+With the initial conditions in equation \ref{eq: van initial} we get $A_1 =
+A_2$ and thus the solution to the oscillation equation is
+\begin{align}
+ I(t) = I_0\cos(\frac{t}{\sqrt{LC}})
+\end{align}
+Now that we know the reduced problem and the solution to it, we may work with
+the Van-Der-Pol equation \ref{eq: vanderpol}, by determining all possible
+non-dimensionalisations. Let us begin by setting
+\begin{align}
+ I(t) = I_c\psi,\\
+ t = t_c \tau,
+\end{align}
+where $I_c$ and $t_c$ have the dimension of $I(t)$ and $t$ accordingly and
+$\psi(\tau)$ and $\tau$ are dimensionless
+The first and second derivative in time is
+\begin{align}
+ \frac{d}{dt} &= \frac{1}{t_c}\frac{d}{d\tau}\\
+ \frac{d^2}{dt^2} &= \frac{1}{t_c^2}\frac{d^2}{d\tau^2}.
+\end{align}
+We can rewrite the Van-Der-Pol equation in terms of $\psi$ and $\tau$
+\begin{align}
+ &\frac{LC}{t_c^2}\ddot{\psi} - \frac{g_1C}{t_c}\dot{\psi}
+ \frac{3g_3CI_c}{t_c}\dot{\psi}\psi = -\psi\\
+ &\psi(0) = \frac{I_0}{I_c} \;\;\;\; \dot{\psi}(0) = 0
+\end{align}
+There are a total of four constants that we can eliminate
+\begin{align}
+ \Pi_1 &= \frac{I_0}{I_c},\;\;\;\;
+ \Pi_2 = \frac{LC}{t_c^2},\nonumber\\
+ \Pi_3 &= \frac{-g_1C}{t_c},\;\;\;\;\;
+ \Pi_4 = \frac{3g_3CI_C}{t_c}.
+\end{align}
+The first scaling is
+\begin{align}
+ I_c = I_0,\;\;\; t_c=\frac{1}{\sqrt{LC}}.
+\end{align}
+Thereby we get the following problem
+\begin{align}
+ &\ddot{\psi}
+ -\sqrt{\frac{C}{L}}g_1\dot{\psi}+\sqrt{\frac{C}{L}}3g_3I_0\dot{\psi}\psi
+ = -\psi\\
+ &\psi(0) = 1 \;\;\;\; \dot{\psi}(0) = 0
+\end{align}
+
+
+The second scaling is
+\begin{align}
+ I_c = I_0,\;\;\; t_c=g_1C.
+\end{align}
+Thereby we get the following problem
+\begin{align}
+ &\frac{L}{g_1^2C}\ddot{\psi}
+ -\dot{\psi}+\frac{3g_3}{g_1}\dot{\psi}\psi
+ = -\psi\\
+ &\psi(0) = 1 \;\;\;\; \dot{\psi}(0) = 0
+\end{align}
+
+The third scaling is
+\begin{align}
+ I_c = I_0,\;\;\; t_c=g_3CI_0.
+\end{align}
+Thereby we get the following problem
+\begin{align}
+ &\frac{L}{g_3^2CI_0^2}\ddot{\psi}
+ -\frac{g_1}{g_3I_0}\dot{\psi}+3\dot{\psi}\psi
+ = -\psi\\
+ &\psi(0) = 1 \;\;\;\; \dot{\psi}(0) = 0
+\end{align}
+\subsection{Scale the Schrödinger Equation}
+The well known Schrödinger equation that describes quantum physics of one
+particle can be written as
+\begin{align}
+ &i\hbar \partial_t\psi = -\frac{\hbar}{2m}\Delta \psi + V\psi \nonumber\\
+ &\psi(t=0) =\psi_0
+\end{align}
+where $\hbar$ is the Planks constant, $\psi=\psi(x, t)$ the wave function,
+$m$ the mass and $V = V(x)$ the potential in which the wave function is. The
+dimensions are
+\begin{align}
+ [\hbar] = Js, \;\;\;\; V = J, \;\;\;\; [\psi]= m^{-d/2}
+\end{align}
+for the special dimension $d$. The standard scaling ansatz is
+\begin{align}
+ &\psi = \psi_c \phi \\
+ &t = t_c \tau \;\;\;\; x = x_c \xi,
+\end{align}
+by that we get the following derivatives in time and in space
+\begin{align}
+ \partial_{x_i} &=\frac{1}{x_{(i)c}} \partial_{\psi_i} \\
+ \partial^2_{x_i} &=\frac{1}{x_{(i)c}^2} \partial_{\psi_i}^2\\
+ \partial_{t} &=\frac{1}{t_c} \partial_{\psi_i} \\
+\end{align}
+for $i = 1, 2, 3$, or depending on the dimension we are dealing with.
+
+Let us consider $x\in \mathbb{R}^3$ and $V = 0$ to scale the equation. First
+we now have
+\begin{align}
+ i \partial_t\psi = -\frac{\hbar}{2m}\Delta \psi
+\end{align}
+with the initial condition $\phi(0) = \frac{\psi_0}{\psi_c}$. With our scaling the equation turns out to be
+\begin{align}
+ \frac{i\hbar t_c}{2m}\frac{1}{||\vec{x}_c||^2}\Delta_{\vec{\xi}}\ \phi =
+ \partial_\tau\ \phi.
+\end{align}
+The constants we get are
+\begin{align}
+ \Pi_1 = \frac{t_c\hbar}{2m}\frac{1}{||\vec{x}_c||^2}, \;\;\;\; \Pi_2 =
+ \frac{\psi_0}{\psi_c}.
+\end{align}
+
+The simple choice of
+\begin{align}
+ \frac{1}{||\vec{x}_c||^2} = 1, \;\;\;\; \psi_c = \psi_0, \;\;\;\; t_c =
+ \frac{2m}{\hbar}||\vec{x}_c||^2,
+\end{align}
+simplifies the Schrodinger equation without the potential to
+\begin{align}
+ i\Delta_{\vec{\xi}}\ \phi = \partial_\tau \phi,
+\end{align}
+with the initial condition $\phi(\tau=0) = 1$.
+
+Now consider $V = 0$, $x\in[0, L]$ and $t \in [0, T]$, the Schrodinger
+equation is the same only with one spacial dimension as above, we can set
+\begin{align}
+ \psi_c = \psi_0, \;\;\;\; x_c =L, \;\;\;\; t_c = \frac{2mL^2}{\hbar}.
+\end{align}
+Thus we get
+\begin{align}
+ i\partial_{\xi}^2 \phi = \partial_\tau \phi,
+\end{align}
+with the initial condition $\phi(\tau=0) = 1$.
+
+As a last example let us consider the quantum harmonic oscillator, that is
+$V(x) = m\omega^2 x^2$ for $x\in \mathbb{R}$, where $\omega$ is the
+frequency, the equation is the following
+\begin{align}
+ i \hbar \partial_t \psi = -\frac{\hbar^2}{2m}\partial^2_x \psi
+ m\omega^2x^2 \psi.
+\end{align}
+By inserting the scaling
+\begin{align}
+ i\partial_\tau \phi = -\frac{t_c\hbar}{2mx_c^2}\partial_\xi^2 \phi
+ \frac{t_cm\omega^2x_c^2}{\hbar} \xi^2 \phi
+\end{align}
+The dimensional constants are
+\begin{align}
+ \Pi_1 = \frac{t_0\hbar}{mx_c^2},\;\;\;\;\Pi_2 =
+ \frac{m\omega^2x_c^2t_c}{\hbar},\;\;\;\; \Pi_3 = \frac{\psi_0}{\psi_c}.
+\end{align}
+The choice of scaling is
+\begin{align}
+ \psi_c = \psi_0, \;\;\;\; t_c = \frac{1}{\omega}, \;\;\;\; x_c =
+ \sqrt{\frac{\hbar}{m\omega}}.
+\end{align}
+Thereby getting the following problem
+\begin{align}
+ i\partial_\tau \phi = -\frac{1}{2} \partial_\xi^2 \phi +\xi^2 \phi
+\end{align}
+with $\phi(\tau = 0) = 1$.
+
+
+
+%\printbibliography
+\end{document}
diff --git a/appl_ana/sesh2/main.pdf b/appl_ana/sesh2/main.pdf
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diff --git a/appl_ana/sesh2/main.tex b/appl_ana/sesh2/main.tex
@@ -0,0 +1,303 @@
+\documentclass[a4paper]{article}
+
+
+\usepackage[T1]{fontenc}
+\usepackage[utf8]{inputenc}
+\usepackage{mlmodern}
+
+%\usepackage{ngerman} % Sprachanpassung Deutsch
+
+\usepackage{graphicx}
+\usepackage{geometry}
+\geometry{a4paper, top=15mm}
+
+\usepackage{subcaption}
+\usepackage[shortlabels]{enumitem}
+\usepackage{amssymb}
+\usepackage{amsthm}
+\usepackage{mathtools}
+\usepackage{braket}
+\usepackage{bbm}
+\usepackage{graphicx}
+\usepackage{float}
+\usepackage{yhmath}
+\usepackage{tikz}
+\usetikzlibrary{patterns,decorations.pathmorphing,positioning}
+\usetikzlibrary{calc,decorations.markings}
+
+%\usepackage[backend=biber, sorting=none]{biblatex}
+%\addbibresource{uni.bib}
+
+\usepackage[framemethod=TikZ]{mdframed}
+
+\tikzstyle{titlered} =
+ [draw=black, thick, fill=white,%
+ text=black, rectangle,
+ right, minimum height=.7cm]
+
+
+\usepackage[colorlinks=true,naturalnames=true,plainpages=false,pdfpagelabels=true]{hyperref}
+\usepackage[parfill]{parskip}
+\usepackage{lipsum}
+
+
+\usepackage{tcolorbox}
+\tcbuselibrary{skins,breakable}
+\newcommand{\eps}{\varepsilon}
+\pagestyle{myheadings}
+
+\markright{Popović\hfill Applied Analysis\hfill}
+
+
+
+\title{Applied Analysis Problems}
+\author{Milutin Popović}
+
+\begin{document}
+\maketitle
+\tableofcontents
+
+\section{Sheet 2}
+\subsection{Problem 4}
+We consider a quadratic equation with two ways to perturb it by $\eps =
+0.001$ \begin{align}
+ x^2 + 2\eps x -1 = 0, \label{eq: (1)}\\
+ \nonumber \\
+ \eps x^2 + 2x - 1 = 0.\label{eq: (2)}
+\end{align}
+Equation \ref{eq: (2)} is singular, because the reduced problem ($\eps
+\rightarrow 0$) has only one solution while the at $x = \frac{1}{2}$. While
+the reduced problem in \ref{eq: (1)} has two solutions for $x = \pm 1$, which
+is the regular case. Let us thereby calculate the asymptotic expansion of the
+regular case up to $O(\eps^2)$, we take the ansatz for the asymptotic
+expansion
+\begin{align}\label{eq: p4 ansatz}
+ x_\eps = x_0 + \eps x_1 + \eps^2 x_2 + O(\eps^3).
+\end{align}
+By substituting $x_\eps$ into \ref{eq: (1)} and factoring out the orders of
+$\eps$ we get
+\begin{align}
+ \eps^0 (x_0^2 - 1) + \eps^1(2x_0 + 2x_0x_1) + \eps^2(x_1^2+2x_2x_0
+ 2x_1) = O(\eps^3)
+\end{align}
+By solving the equations in order of $\eps$, for the coefficients
+$x_0$, $x_1$ and $x_2$ we get
+\begin{align}
+ x_0 = \pm 1, \;\;\;\; x_1 = -1, \;\;\;\; x_2 = \pm \frac{1}{2}.
+\end{align}
+By substituting into the equation \ref{eq: p4 ansatz} we get
+\begin{align}
+ x_\eps = \pm 1 - \eps \pm \frac{1}{2} \eps + O(\eps^3).
+\end{align}
+For $\eps = 0.001$ we get
+\begin{align}
+ x_\eps = -1.00150 + O(\eps^3)\\
+ x_\eps = -1.00100 + O(\eps^2).
+\end{align}
+\subsection{Problem 5}
+Consider the following equations
+\begin{align}
+ \label{eq: p5 1}\eps y' + y = x \;\;\;\;\;\; y(0) = 1\\
+ \label{eq: p5 2}\eps y' + y = x \;\;\;\;\;\; y(0) = 0\\
+ \nonumber\\
+ \label{eq: p5 3}\eps y' + y = x \;\;\;\;\;\; y(0) = \eps\\
+ \label{eq: p5 4}\eps^2 y' + y = x \;\;\;\;\;\; y(0) = \eps\\
+ \nonumber\\
+ \label{eq: p5 5}y' + \eps y = x \;\;\;\;\;\; y(0) = 1\\
+ \label{eq: p5 6}y' + y = \eps x \;\;\;\;\;\; y(0) = 1\\
+\end{align}
+We will go through the equations and elaborate on if the perturbation is
+regular or singular, if regular we will compute the asymptotic expansion up
+to second order.
+Let us begin with equation \ref{eq: p5 1}. By the first look, the reduced
+problem does not agree with the boundary condition
+\begin{align}
+ y_0 = x \;\;\;\;\; y_0(0) = 1,
+\end{align}
+is a contradiction in $y_0(0) = 0 \neq 1$, thereby equation \ref{eq: p5 1} is
+\textbf{singularly perturbed}.
+
+The reduced problem of equation \ref{eq: p5 2} on the other hand agrees with
+the boundary condition, since
+\begin{align}
+ y_0 = x \;\;\;\;\; y_0(0) = 0.
+\end{align}
+But by doing the ansatz for the asymptotic expansion
+\begin{align}
+ y_\eps(x) = y_0 + \eps y_1 + \eps^2 y_2 + O(\eps^3),
+\end{align}
+plugging in into \ref{eq: p5 2} and separating coefficients in terms of
+$\eps$, we get
+\begin{align}
+ \eps^0 (y_0 -x) + \eps^1(y_0' + y_1) + \eps^2(y_1' + y_2) = O(\eps^3)
+\end{align}
+The solutions to these equations are
+\begin{align}
+ y_0 = x, \;\;\;\; y_1 = 1, \;\;\;\; y_2 = 0,
+\end{align}
+which is a contradiction to the boundary condition of $y_1(0) = 1 \neq 0$.
+Thereby we can conclude that equation \ref{eq: p5 2} is \textbf{singularly
+perturbed}.
+
+Next up is equation \ref{eq: p5 3}, where by the asymptotic expansion the
+first order coefficient of $\eps$, $y_2$, has the boundary condition $y_2(0)
+= 0$. But by applying the ansatz of the asymptotic expansion and plugging
+into the equation we get
+\begin{align}
+ \eps^2(y_0 - x) + \eps^1 (y_0' + y_1) + \eps^2(y_1' + y_2) = O(\eps^3).
+\end{align}
+Solving these equations we get
+\begin{align}
+ y_0 = 0, \;\;\;\; y_1 = 1 \;\;\;\; y_2 = 0 ,
+\end{align}
+which is a contradiction $y_1(0) = 1 \neq 0 $, thus the equation \ref{eq: p5
+3} is \textbf{singularly perturbed}.
+
+The next equation \ref{eq: p5 4} is also singularly perturbed, we
+can see this by plugging the asymptotic expansion into the equation
+\begin{align}
+ \eps^0 ( y_0 - x) + \eps^1(y_1) + \eps^2(y_0' + y_2) = O(\eps^3),
+\end{align}
+solving for the coefficients we get
+\begin{align}
+ y_0 = x, \;\;\;\; y_1 = 0, \;\;\;\;\; y_2 = -1,
+\end{align}
+which is contradiction by the boundary condition $y_2(0) = -1 \neq 0$,
+thereby \ref{eq: p5 4} is \textbf{singularly perturbed}.
+
+Equation \ref{eq: p5 5} on the first sight does not indicate for any
+contradictions, we may plug the ansatz of the asymptotic expansion into the
+equation and see what happens
+\begin{align}
+ \eps^0(y_0 -x) + \eps^1(y_1' + y_0) + \eps^2(y_2' +y_1) = O(\eps^2),
+\end{align}
+with the initial conditions $y_0(0) = 1$, $y_1(0) = y_2(0) = 0$.
+\begin{align}
+ y_0 = \frac{x^2}{2} + 1, \;\;\;\;
+ y_1 = -\frac{x^3}{6} + x, \;\;\;\;
+ y_2 = \frac{x^4}{24} + \frac{x^2}{2}.
+\end{align}
+Finally we get
+\begin{align}
+ y_\eps(x) = (\frac{x^2}{2}+1) + \eps(-\frac{x^3}{6} -x)
+ +\eps^2(\frac{x^4}{24} + \frac{x^2}{2}) + O(\eps^3).
+\end{align}
+Thereby we can conclude that \ref{eq: p5 5} is \textbf{regularly perturbed}.
+
+The last equation \ref{eq: p5 6} is also regular, let us do the asymptotic
+expansion of the equation and order the equation in orders of $\eps$.
+\begin{align}
+ \eps^0(y_0' + y_0) + \eps^1(y_1' + y_1 -x) + \eps^2(y_2 + y_2) =
+ O(\eps^3).
+\end{align}
+by solving these differential equations with the boundary conditions $y_0(0)
+= 1$, $y_1(0) = y_2(0) = 0$ we get
+\begin{align}
+ y_0 = e^{-x} \;\;\;\; y_1 = (x-1) + e^{-x} \;\;\;\; y_2 = 0.
+\end{align}
+The equation we get
+\begin{align}
+ y_\eps(x) = e^{-x} + \eps(x-1 +e^{-x}) + O(\eps^3).
+\end{align}
+Thereby we can conclude that the last equation \ref{eq: p5 6} is
+\textbf{regularly perturbed}.
+\subsection{Problem 6}
+In this section we will calculate the asymptotic expansion of a regularly
+perturbed equation in two ways, by doing the regular expansion ansatz and by
+substituting and expanding in terms of $\eps$. The ordinary differential
+equation we are dealing with is
+\begin{align}
+ y' = -y + \eps y^2 \;\;\;\;\; y(0) = 1,
+\end{align}
+where $t > 0$ and $0 < \eps \ll 1$. The standard expansion ansatz is
+\begin{align}
+y_\eps(x) = y_0 + \eps y_1 + \eps^2 y_2 + O(\eps^3).
+\end{align}
+The ODE then expands to
+\begin{align}
+ \eps^0(y_0' + y_0) + \eps(y_1' + y_1 - y_0^2) + \eps^2(y_2' + y_2 -
+ 2y_0y_1) = O(\eps^3).
+\end{align}
+Equations in order of $\eps$ and $\eps^2$ are non homogeneous ODE's, a little
+bit of a hassle to solve but, still easy by first solving the non homogeneous
+part, getting a constant dependent on $x$ we can plug the solution into the
+non homogeneous equation and calculate for the constant (...). The solution to
+these three coefficients with the boundary conditions $y_0(0) = 1$, $y_1(0) =
+y_2(0) = 1$ we get
+\begin{align}
+ y_0 = e^{-x}, \;\;\;\; y_1 = -e^{-2x} + e^{-x}, \;\;\;\; y_2 = e^{-3x} -
+ 2e^{-2x} + e^{-x}.
+\end{align}
+The expansion of $y$ is then
+\begin{align}
+ y_\eps (x) = e^{-x} + \eps(-e^{-2x} + e^{-x}) + \eps^2(e^{-3x} - 2e^{-2x}
++ e^{-x}) + O(\eps^3). \end{align}
+
+The second ansatz, considers the substitution $z = \frac{1}{y}$, by
+calculating the first derivative and substituting the original problem we
+get
+\begin{align}
+ z' &= \frac{-y'}{y^2} = \frac{y-\eps y^2}{y^2} = \frac{1}{y} - \eps = z -
+ \eps. \\
+ z(0) &= \frac{1}{y(0)} = 1.
+\end{align}
+The solution is
+\begin{align}
+ z(x) = \eps + (1-\eps) e^x.
+\end{align}
+By substituting this into $y = \frac{1}{z}$ and expanding we get
+\begin{align}
+ y(x) &= \frac{1}{\eps+(1-\eps)e^x} = e^{-x} \frac{1}{1 - (1- e^{-x})\eps}
+ \\
+ &= e^{-x} \sum_{n\geq 0} \eps^n(1-e^{-x})^n.
+\end{align}
+which is the geometric series.
+\subsection{Problem 7}
+The last problem consists of a perturbation of a partial differential
+equation (heat equation).
+\begin{align}
+ &\partial_t u(x, t) + \partial_x^2 u(x,t) - \eps u(x, t)^2 = 0
+ &x\in (0, 1),\; t>0,\\
+ &u(x, 0) = \tilde{u}_0(x) &x\in(0, 1), \\
+ &u(0, t) = u(1, t) = 0 & t>0.
+\end{align}
+The problem is regular because the reduced solution is the regular heat
+equation in the one special dimension on $x\in (0, 1)$, we know this is
+solvable. By doing the expansion ansatz we can derive the first equations
+for the first three terms, the ansatz is always the same
+\begin{align}
+ u_\eps = u_0 + \eps u_1 + \eps^2 u_2 + O(\eps^3).
+\end{align}
+Plugging this into the perturbed problem problem and factoring out the terms
+in the order of $\eps$ we get
+\begin{align}
+ &\eps^0 (\partial_t u_0 + \partial_x^2 u_0) + \\
+ &\eps^1 (\partial_t u_1 + \partial_x^2 u_1 - u_0^2) +\\
+ &\eps^2 (\partial_t u_2 + \partial_x^2 u_2 - 2u_1u_0) = O(\eps^3)
+\end{align}
+
+We can solve the reduced problem with the initial condition $\tilde{u}_0 =
+\sin(\pi x)$ by separation of variables. Setting $u(x, t) = \xi(x) \tau(t)$
+and substituting into the equation we get two ordinary differential equation
+\begin{align}
+ \underbrace{\frac{\xi_{xx}}{\xi}}_{=k}
+ \underbrace{\frac{\tau_t}{\tau}}_{=-k} = 0
+\end{align}
+for some $k$. Solving these two by the exponential ansatz.
+\begin{align}
+ \xi(x) &= A_1 e^{\sqrt{k} x} +A_2 e^{-\sqrt{k} x},\\
+ \tau(t) &= A_3 e^{-kt}.
+\end{align}
+With the initial condition we get the conditions that
+\begin{align}
+ A_1A_3 &= -A_2 A_3,\\
+ k &= \pi^2
+\end{align}
+we choose $A_1 = A_3 = 1$ , $A_2 = -1$. We get the following solution to the
+PDE
+\begin{align}
+ u(x, t) = \xi(x)\tau(t) = \sin(\pi x) e^{-\pi^2 t}.
+\end{align}
+
+%\printbibliography
+\end{document}
