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Sharedlatex-tests / long-presentation.texOpen in CoCalc
Author: Harald Schilly
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\usetheme{Madrid}%CambridgeUS}%CambridgeUS}%Hannover}%default}%AnnArbor}%Goettingen}%Antibes}%CambridgeUS} %Madrid} %Frankfurt}%Warsaw}
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\title[Perturbation Theory]{Introduction to Ordinary Differential Equations}
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\author[Emily Weymier]{Emily Weymier}
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\institute[SFA]{\begin{tabular}{l} Department of Mathematics $\&$ Statistics \\Stephen F. Austin State University, Nacogdoches, TX
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\end{tabular}}
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\date{September 22, 2017}
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\begin{document}
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\frame{\titlepage}
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\begin{frame}{Outline}
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\begin{enumerate}
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\item What is a differential equation? \vfill
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\item Initial Value Problems %\vfill
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\begin{itemize}
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\item Linear first order differential equations
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\item Second order differential equations
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\item Recasting high order differential equations as a system of first order differential equations
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\end{itemize} \vfill
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\item Boundary Value Problems \vfill
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\item Solution techniques for nonlinear differential equations %\vfill
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\begin{itemize}
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\item Power series solutions
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\item Perturbation theory concept
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\end{itemize} \vfill
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\item Concluding Remarks \vfill
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\end{enumerate}
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\end{frame}
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\section{What is a differential equation?}
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\begin{frame}{Differential Equations: The Basics}
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\begin{itemize}
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\item Ordinary differential equations are used to model change over an independent variable (for our purposes it will usually be $t$ for time) without using partial derivatives. \pause
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\vfill
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\item Differential equations contain three types of variables: an independent variable, at least one dependent variable (these will be functions of the independent variable), and the parameters.\pause
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\vfill
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\item ODE's can contain multiple iterations of derivatives. They are named accordingly (i.e. if there are only first derivatives, then the ODE is called a first order ODE).\pause
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\vfill
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%\item
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%\vfill
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\end{itemize}
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\end{frame}
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\begin{frame}{A Simple Example: Population Modeling}
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Population growth is commonly modeled with differential equations. In the following equation: $t=$ time, $P=$ population and $k=$ proportionality constant. $k$ represents the constant ratio between the growth rate of the population and the size of the population.
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\begin{eqnarray*}
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\frac{dP}{dt} = kP
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\end{eqnarray*}
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In this particular equation, the left hand side represents the growth rate of the population being proportional to the size of the population $P$. This is a very simple example of a first order, ordinary differential equation. The equation only contains first order derivatives and there are no partial derivatives. %I think it is slipping my mind right now, but is there a word for non partial derivatives? That is probably a stupid question but I am blanking on it right now.
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\end{frame}
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\section{Initial Value Problems}
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\begin{frame}{Initial Value Problems}
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An initial value problem consists of a differential equation and an initial condition. So, going back to the population example, the following is an example of an initial value problem:
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\begin{eqnarray*}
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\frac{dP}{dt} = kP, P(0)=P_0
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\end{eqnarray*}
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The solution to this set of equations is a function, call it $P(t)$, that satisfies both equations.
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\end{frame}
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\begin{frame} {Linear First Order Differential Equations}
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\begin{itemize}
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\item The standard form for a first-order differential equation is
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\begin{eqnarray*}
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\frac{dy}{dt} = f(t,y)
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\end{eqnarray*}
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where the right hand side represents the function $f$ that depends on the independent variable, $t$, and the dependent variable, $y$.
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\end{itemize}
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\end{frame}
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\begin{frame}{General Solutions to a Differential Equation}
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Let's look at a simple example and walk through the steps of finding a general solution to the following equation
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\begin{eqnarray*}
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\frac{dy}{dt} = (ty)^2
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\end{eqnarray*}
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\vfill
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We will simply "separate" {\color{red}write as ``separate''} the variables then integrate the both sides of the equation to find the general solution.
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\begin{eqnarray*}
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\frac{dy}{dt} &=& t^2 y^2
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\\ \frac{1}{y^2} \,dy &=& t^2 \, dt
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\\ \int \frac{1}{y^2} \,dy &=& \int t^2 \, dt
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\end{eqnarray*}
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\end{frame}
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\begin{frame}
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\begin{eqnarray*}
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\\ - y^{-1} &=& \frac{t^3}{3} + c
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\\ -\frac{1}{y} &=& \frac{t^3}{3} + c
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\\ y &=& -\frac{1}{\frac{t^3}{3} + c}
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\\ \Rightarrow y(t) &=& -\frac{3}{t^3 + c_1}
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\end{eqnarray*}
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where $c_1$ is any real number.
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\end{frame}
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\section{Solving Initial Value Problems}
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\begin{frame}{Linear First Order Differential Equations}
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Initial value problems consist of a differential equation and an initial value. We will work through the example below:
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\begin{eqnarray*}
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\frac{dx}{dt} = -xt;~~~~~ x(0) = \frac{1}{\sqrt{\pi}}
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\end{eqnarray*}
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First we will need to find the general solution to $\frac{dx}{dt} = -xt$, then use the initial value $x(0)=\frac{1}{\sqrt{\pi}}$ to solve for $c$. Since we do not know what $x(t)$ is, we will need to "separate" the equation before integrating.
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\begin{eqnarray*}
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\frac{dx}{dt} &=& -x t
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\\ -\frac{1}{x} \,dx &=& t \,dt % "separate" the variables
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\\ \int -\frac{1}{x} \,dx &=& \int t \,dt % integrate both sides with respect to the appropriate variable
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\end{eqnarray*}
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\end{frame}
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\begin{frame}{Linear First Order Differential Equations Continued}
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\begin{eqnarray*}
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-\ln{x} &=& \frac{t^2}{2} + c
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\\ x &=& e^{-(\frac{t^2}{2} + c)}
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\\ x &=& e^{-(\frac{t^2}{2})} e^{-c}
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\\ x &=& k e^{-\frac{t^2}{2}}
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\end{eqnarray*}
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The above function of $t$ is the general solution to $\frac{dx}{dt} = -xt$ where $k$ is some constant. Since we have the initial value $x(0) = \frac{1}{\sqrt{\pi}}$, we can solve for $k$.
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\end{frame}
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\begin{frame}{Solving Initial Value Problems}
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Thus we can see that the solution to the initial value problem
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\begin{eqnarray*}
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\frac{dx}{dt} = -xt; x(0) = \frac{1}{\sqrt{\pi}}
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\end{eqnarray*}
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is
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$$x(0) = \frac{1}{\sqrt{\pi}} = k e^{-\frac{0^2}{2}}$$
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$$x(t) = \frac{1}{\sqrt{\pi}} \, e^{-\frac{t^2}{2}}$$ %is this even right? OK I think I got it this time.
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\end{frame}
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\begin{frame}
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Let's verify that this solution is correct. We will need to show
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\begin{eqnarray*}
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\frac{dx}{dt} &=& x'(t) = f(t, x(t)) %I don't feel like this is the correct notation.??
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\\\frac{dx}{dt} &=& \frac{d}{dt}\bigg(\frac{1}{\sqrt{\pi}} e^{-\frac{t^2}{2}} \bigg) = \frac{1}{\sqrt{\pi}} e^{-\frac{t^2}{2}}
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\\ &\Rightarrow& \frac{d}{dt}\bigg(\frac{1}{\sqrt{\pi}} e^{-\frac{t^2}{2}} \bigg) = -\frac{1}{\sqrt{\pi}} e^{-\frac{t^2}{2}}
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\end{eqnarray*}
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\end{frame}
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\begin{frame}{Second Order Differential Equations}
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Second order differential equations simply have a second derivative of the dependent variable. The following is a common example that models a simple harmonic oscillator:
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\begin{eqnarray*}
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\frac{d^2y}{dt^2} + \frac{k}{m} y = 0
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\end{eqnarray*}
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where $m$ and $k$ are determined by the mass and spring involved. This second order differential equation can be rewritten as the following first order differential equation:
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\begin{eqnarray*}
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\frac{dv}{dt} = -\frac{k}{m} y
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\end{eqnarray*}
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where $v$ denotes velocity.
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\end{frame}
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\begin{frame}{Second Order Differential Equations Continued}
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Referring back to some calculus knowledge, if $v(t)$ is velocity, then $v=\frac{dy}{dt}$. Thus, we can substitute in $\frac{dv}{dt}$ into our second order differential equation and essentially turn it into a first order differential equation.
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\begin{eqnarray*}
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\frac{d^2y}{dt^2} = -\frac{k}{m} y \Leftrightarrow \frac{dv}{dt} = -\frac{k}{m} y
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\end{eqnarray*}
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Now we have the following system of first order differential equations to describe the original second order differential equation:
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\begin{eqnarray*}
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\frac{dy}{dt} &=& v
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\\ \frac{dv}{dt} &=& -\frac{k}{m} y
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\end{eqnarray*}
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\end{frame}
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\begin{frame}{Second Order Differential Equations Continued}
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Consider the following initial value problem:
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$$\frac{d^2y}{dt^2} + y = 0$$
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with $y(0) = 0$ and $y'(0) = v(0) = 1$. Let's show that $y(t) = \sin(t)$ is a solution. Let $v=\frac{dy}{dt}$, then we have the following system:
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\begin{eqnarray*}
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\frac{dy}{dt} &=& v
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\\ \frac{dv}{dt} &=& -y
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\end{eqnarray*}
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\end{frame}
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\begin{frame}{Second Order Differential Equations Continued}
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\begin{eqnarray*}
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\frac{dy}{dt} &=& \frac{d}{dt} \sin(t) = \cos(t) = v
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\\ \frac{dv}{dt} &=& -\sin(t) = -y
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\\ \Rightarrow \frac{d^2y}{dt^2} &=& -\sin(t)
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\\ \Rightarrow \frac{d^2y}{dt^2} + y &=& \frac{d^2 (\sin(t))}{dt^2} + \sin(t)
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\\ &=& -\sin(t) + \sin(t) = 0
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\end{eqnarray*}
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\end{frame}
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\begin{frame}{High Order Differential Equations as a System}
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\end{frame}
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\section{Boundary Value Problems}
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\begin{frame}{Boundary Value Problems: The Basics}
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\end{frame}
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\section{Solution Techniques for Nonlinear Differential Equations}
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\begin{frame}{Power Series Solutions}
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To demonstrate how to use power series to solve a nonlinear differential equation we will look at Hermite's Equation:
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\begin{eqnarray*}
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\frac{d^2y}{dt^2} - 2t \frac{dy}{dt} + 2 py = 0
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\end{eqnarray*}
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We will use the following power series and its first and second derivatives to make a guess:
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\begin{eqnarray}
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\label{function} y(t) &=& a_0 + a_1 t + a_2 t^2 + a_3 t^3 + ... = \sum_{n=0}^{\infty} a_n t^n
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\\ \label{der} \frac{dy}{dt} &=& a_1 + 2a_2 t + 3a_3 t^2 + 4a_4 t^3 + ... = \sum_{n=1}^{\infty} n a_n t^{n-1}
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\\ \label{der2} \frac{d^2y}{dt^2} &=& 2 a_2 + 6 a_3 t + 12 a_4 t^2 + ... = \sum_{n=2}^{\infty} n(n-1) a_n t^{n-2}
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\end{eqnarray}
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\end{frame}
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\begin{frame}
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From the previous equations we can conclude that
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\begin{eqnarray*}
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y(0) &=& a_0
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\\ y'(0) &=& a_1
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\end{eqnarray*}
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Next we will substitute (\ref{function}), (\ref{der}) and (\ref{der2}) into Hermite's Equation and collect like terms.
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\begin{eqnarray*}
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\frac{d^2y}{dt^2} - 2t \frac{dy}{dt} + 2 py = 0 = (2 a_2 + 6 a_3 t + 12 a_4 t^2 + ...)
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\\ - 2t (a_1 + 2a_2 t + 3a_3 t^2 + 4a_4 t^3 + ...)
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\\ +2p (a_0 + a_1 t + a_2 t^2 + a_3 t^3 + ...)
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\\ \Rightarrow (2pa_0 + 2a_2) + (2pa_1 - 2a_1 + 6a_3)t +
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\\ (2pa_2 - 4a_2 + 12a_4)t^2 + (2pa_3 - 6a_3 + 20a_5)t^3 = 0
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\end{eqnarray*}
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\end{frame}
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\begin{frame}
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Then from here, we will set all coefficients equal to $0$ since the equation is equal to $0$ and $t \neq 0$. %is that true?
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We get the following sequence of equations:
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\begin{eqnarray*}
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2pa_0 + 2a_2 &=& 0
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\\ 2pa_1 - 2a_1 + 6a_3 &=& 0
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\\ 2pa_2 - 4a_2 + 12a_4 &=& 0
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\\ 2pa_3 - 6a_3 + 20a_5 &=& 0
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\end{eqnarray*}
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\end{frame}
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\begin{frame}
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Then will several substitutions we arrive at the following set of equations:
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\begin{eqnarray*}
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\\ \Rightarrow a_2 &=& -pa_0
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\\ a_3 &=& - \frac{p-1}{3} a_1
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\\ a_4 &=& -\frac{p-2}{6} a_2 = \frac{(p-2)p}{6} a_0
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\\ a_5 &=& -\frac{p-3}{10} a_3 = \frac{(p-3)(p-1)}{30} a_1
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\end{eqnarray*}
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\end{frame}
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\begin{frame}{Perturbation Theory Concept}
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Perturbation theory is used when a mathematical equation involves a small perturbation, usually $\epsilon$. From here we create $y(x)$ such that it is an expansion in terms of $\epsilon$. For example
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\begin{eqnarray*}
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y(x) = y_0(x) + \epsilon y_1(x) + \epsilon^2 y_2(x)+\cdots
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\end{eqnarray*}
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This summation is called a perturbation series and it has a nice feature that allows each $y_i$ to be solved using the previous $y_i$'s.
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Consider the equation,
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\begin{eqnarray}
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\label{ex1} x^2+x+6\epsilon &=& 0, \hspace{.5cm} \epsilon \ll 1
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\end{eqnarray}
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\noindent Let's consider using perturbation theory to determine approximations for the roots of Equation (\ref{ex1}).
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\end{frame}
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\begin{frame}{Perturbation Theory Concept Continued}
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Notice this equation is a perturbation of $x^2+x=0$. Let $x(\epsilon) = \sum_{n=0}^{\infty} a_n \epsilon^n$. This series will be substituted into (\ref{ex1}) and powers of $\epsilon$ will be collected. %We wish for the equation to be true for any $\epsilon \ll 1$.
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Next we will calculate the first term of the series by setting $\epsilon =0$ in (\ref{ex1}). So the leading order equation is
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\begin{eqnarray}
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\label{LOE} a_0^2 + a_0 = 0
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\end{eqnarray}
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\noindent with solutions $x = -1, 0$. Thus $x(0) = a_0 = -1, 0$.
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Now the perturbation series are as follows
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\begin{eqnarray}
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&=& 1 - a_1 \epsilon - a_2 \epsilon^2 - a_1 \epsilon + a_1^2 \epsilon^2 + a_1 a_2 \epsilon^3 - a_2 \epsilon^2 + a_1 a_2 \epsilon^3 + a_2^2 \epsilon^4 - 1 + a_1 \epsilon + a_2 \epsilon^2 + 6 \epsilon
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%\\ = 1 - 2 a_1 \epsilon - 2 a_2 \epsilon^2 + a_1^2 \epsilon^2 + 2 a_1 a_2 \epsilon^3 + a_2^2 \epsilon^4 - 1 + a_1 \epsilon + a_2 \epsilon^2 + 6 \epsilon %combining like terms
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\\ &=& (1-1) + (-2 a_1 + a_1 + 6)\epsilon + (-2 a_2 + a_1^2 +a_2)\epsilon^2 + \mathcal{O}(\epsilon^3) %factoring out epsilons.
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\end{eqnarray}
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\end{frame}
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\begin{frame}{Perturbation Theory Concept Continued}
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\begin{eqnarray}
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\\\label{root1} x_1(\epsilon) = -1 + a_1 \epsilon + a_2 \epsilon^2 + \mathcal{O}(\epsilon^3)
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\end{eqnarray}
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and
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\begin{eqnarray}
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\label{root2} x_2(\epsilon) &=& 0 + b_1 \epsilon + b_2 \epsilon^2 + \mathcal{O}(\epsilon^3)
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\end{eqnarray}
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Next, we will substitute in (\ref{root1}) into (\ref{ex1}) while ignoring powers of $\epsilon$ greater than $2$. Since we are only approximating the solution to the second-order, we can disregard the powers of $\epsilon$ greater than $2$.
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\begin{align*}
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x^2 + x + 6 \epsilon &= (-1 + a_1 \epsilon + a_2 \epsilon^2)^2 + (-1 + a_1 \epsilon + a_2 \epsilon^2) + 6 \epsilon
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\\ &\Rightarrow (- a_1 + 6)\epsilon + (- a_2 + a_1^2)\epsilon^2 + \mathcal{O}(\epsilon^3)
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\end{align*}
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\end{frame}
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\begin{frame}{Perturbation Theory Concept Continued}
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From here we take the coefficient of each power of $\epsilon$ and set it equal to zero. This step is justified because (\ref{ex1}) is equal to zero and $\epsilon \neq 0$ so each coefficient must be equal to zero. Thus we have the following equations
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\begin{align*}
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\mathcal{O}(\epsilon^1) &: -a_1 + 6 = 0
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\\ \mathcal{O}(\epsilon^2) &: a_1^2 - a_2 = 0
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\end{align*}
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These equations will be solved sequentially. The results are $a_1 = 6$ and $a_2 = 36$. Thus the perturbation expansion for the root $x_1 = -1$ is:
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$$x_1(\epsilon) = -1 + 6 \epsilon + 36 \epsilon^2 + \mathcal{O}(\epsilon^3)$$
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The same process can be repeated for $x_2$ with the perturbation expansion for the root $x_2 = 0$ resulting in
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$$x_2(\epsilon) = -6 \epsilon - 36 \epsilon^2 + \mathcal{O}(\epsilon^3)$$
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\end{frame}
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%Here's the tex and work for the above:
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%Now, we will plug equation (\ref{root2}) into (\ref{ex1}) while disregarding powers of $\epsilon$ greater than $2$. This will create the perturbation expansion for the root $x_2 = 0$.
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%\begin{align*}
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%x^2 + x + 6 \epsilon &= (0 + b_1 \epsilon + b_2 \epsilon^2)^2 + (0 + b_1 \epsilon + b_2 \epsilon^2) + 6 \epsilon
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% &= b_1^2 \epsilon^2 + 2 b_1 b_2 \epsilon^3 + b_2^2 \epsilon^4 + b_1 \epsilon + b_2 \epsilon^2 + 6 \epsilon
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%\\ &= (b_1 + 6) \epsilon + (b_1^2 + b_2) \epsilon^2 + \mathcal{O}(\epsilon^3)
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%\end{align*}
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%From here we take the coefficient of each power of $\epsilon$ and set it equal to zero. This step is justified because (\ref{ex1}) is equal to zero and $\epsilon \neq 0$ so each coefficient must be equal to zero. Thus we have the following equations
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%\begin{align*}
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%\mathcal{O}(\epsilon^1) &: b_1 + 6 = 0
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%\\ \mathcal{O}(\epsilon^2) &: b_1^2 + b_2 = 0
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%\end{align*}
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%These equations will be solved sequentially. The results are
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%\begin{align*}
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%b_1 &= -6
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%\\ b_1^2 + b_2 &= 0
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%\\ (-6)^2 + b_2 &= 0
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%\\ b_2 &= -36
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%\end{align*}
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%Thus the perturbation expansion for the root $x_2 = 0$ is
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%$$x_2(\epsilon) = -6 \epsilon - 36 \epsilon^2 + \mathcal{O}(\epsilon^3)$$
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%\begin{frame}{Numerical Experiments}
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%\begin{itemize}
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%\item Example 1 - Test convergence rate, computational efficiency, and accuracy with known exact solution\vfill\pause
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%\item Example 2 - Test convergence rate, computational efficiency, and accuracy with nonlinear model\vfill\pause
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%\item Example 3 - convergence rate, computational efficiency of three-species model\vfill
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%\end{itemize}
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%\end{frame}
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\begin{frame}{Concluding Remarks}
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\end{frame}
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%\begin{frame}{Efficiency}
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%\begin{minipage}{1.0\textwidth}
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%\begin{figure}[h]
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%\begin{center}
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%\includegraphics[scale=.3]{Example1CompTime.pdf}
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%\caption{A log-log plot of the computational time, in seconds, versus $N$ after $1000$ iterations. The temporal step is held constant, $\tau = 10^{-6}$, while $h=1/(N-1)$. A linear least squares approximates the slope of the line to be $1.654628$. This indicates that the computational time is scales as $N^{1.654628}$. The computational time of an efficient scheme should scale no slower than $N^2$. This indicates that the proposed nonlinear splitting scheme is highly efficient.}
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%\end{center}
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%\end{figure}
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%\end{minipage}
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%\begin{minipage}{1.0\textwidth}
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%\begin{itemize}
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%\item Let $\tau = 10^{-6}$ and consider the computational time for $1000$ temporal steps for $N=21,~31, \ldots, 401$.\vfill\pause
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%\item The computational time scales as $N^{1.654628}$\vfill
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%\end{itemize}
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%\end{minipage}
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%\end{frame}
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%\begin{frame}{Example 2 - Invasive Species Model}
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%\end{frame}
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%\begin{frame}{Nonlinear Splitting Algorithm}
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%\end{frame}
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%\begin{frame}{All in the details}
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%\end{frame}
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%\begin{frame}{Concluding Remarks}
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%\begin{itemize}
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%\item In summary:
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% \begin{enumerate}
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% \item \textbf{Positivity} of numerical solution guaranteed.
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% \vfill
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% \item \textbf{Nonlinear} stability and convergence were shown.
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% \vfill
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% \item \textbf{Computationally efficient} design $\rightarrow$ Computational Time scales less than $N^2$.
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% \end{enumerate}
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% \vfill
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%\end{itemize}
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%\end{frame}
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\begin{frame}{Questions?}
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\end{document}
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