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practice_mt1.tex
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practice_mt1.tex
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\documentclass[12pt,letterpaper]{article}
%\usepackage[secthm,mdthm,simplethm]{beel}
\usepackage[utf8]{inputenc}
\usepackage{graphicx, tcolorbox}
\usepackage[margin = 1in, top = 0.8in,bottom = 0.8in]{geometry}
\usepackage{amsmath,amsfonts,amsthm,amssymb}
\usepackage{array,comment,enumitem}
\usepackage{mathtools,thmtools}
\usepackage{multicol,tabto,setspace}
\usepackage{tikz,pgfplots,tikz-cd,venndiagram,forest}
\usepackage[all]{xy}
\usepackage{listings,fancyhdr}
\pagestyle{fancy}
\headheight = 15 pt
\lhead{Bill Li}
\rhead{Partial Differential Equation}
\cfoot{\thepage}
\pgfplotsset{compat = 1.15}
\usetikzlibrary{fadings}
\hbadness = 10000
\tolerance = 10000
\renewcommand{\qedsymbol}{$\blacksquare$}
\renewcommand{\bar}{\overline}
\renewcommand{\b}{\mathbb}
\renewcommand{\c}{\mathcal}
\newcommand{\s}{\mathscr}
\begin{document}
\begin{center}
{\Large \textbf{Unofficial Solution to Practice Problem}}
\end{center}
\begin{enumerate}
\item Find the solution $u = u(x,t)$ to
\[ \left\{ \begin{array}{ll}
u_t + 3xu_x = 0 & \mathrm{for} \quad - \infty < x < \infty, -\infty < t < \infty \\
u(x,0) = \sin x & \mathrm{for} \quad -\infty < x < \infty \\
\end{array} \right. \]
\begin{proof}[Solution]
Observe that the solution of the PDE must be tangent to $(1,3x)$ all the time. Therefore we have
\[ \frac{dx}{dt} = \frac{3x}{1} \]
Solving for the ODE gives us
\[ x = Ce^{3t} \]
Since $ u_t + 3xu_x = 0$, the curve is independent of $t$. Solving for $C$ gives us $C = e^{-3t}x$, therefore we can rewrite $u$ as a function of one variable.
\[u(x,t) = u(Ce^{3t},t) = u(C,0) = f(e^{-3t}x)\]
Now we plug in the initial condition we have
\begin{align*}
u(x,0) = f(e^{-3 \cdot 0} x) = \sin x \implies f(x) = \sin x,
\end{align*}
therefore the solution to the PDE is
\[ u(x,t) = \sin(e^{-3t}x)\]
\end{proof}
\newpage
\item \begin{enumerate}
\item Find the coefficients of the Fourier cosine Series of $f(x) = x$ on $[0,\pi]$.
\begin{proof}
Let $l = \pi$. We know the full Fourier Series of $f(x)$ is of the form
\[ x = \frac 12 A_0 + \sum_{n = 1}^\infty A_n \cos \frac{n \pi x}{l}.\]
We shall compute for $A_0$ first
\[ A_0 = \frac 2l \int_0^l x \cdot 1 \, \mathrm{d}x = \pi.\]
Then for $n \neq 0$ we have
\begin{align*}
A_n &= \frac 2l \int_0^l x \cdot \cos \frac{n \pi x}{l} \, \mathrm{d}x \\
&= \frac 2l \left[ \frac{l}{n \pi} x \cdot \sin \frac{n \pi x}{l} + \frac{l^2}{(n\pi)^2} \cos \frac{n \pi x}{l} \right]_0^l \\
&= \frac 2\pi \left[ \frac 1nx \cdot \sin nx + \frac 1{n^2} \cos nx \right]_0^\pi \\
&= \frac{2}{\pi n^2} \left( \cos n \pi - \cos 0 \right) \\
&= \frac{2}{\pi n^2} ((-1)^n - 1)
\end{align*}
Therefore the Fourier cosine series for $x$ is
\[x = \frac \pi 2 - \sum_{n=1}^{\infty} \frac{4}{2n \pi} \cos n x\]
Therefore we have
\[A_n = \left\{ \begin{array}{ll}
-\frac{4}{n\pi} & n \, \mathrm{odd} \\
0 & n \, \mathrm{even} \\
\end{array} \right. \]
Therefore the Fourier cosine series for $x$ is
\[ x = \frac \pi 2 - \sum_{n=1}^{\infty} \frac{4}{(2n + 1)\pi} \cos(2n +1) x\]
\end{proof}
\item Prove that the cosine series converges uniformly in $[0,\pi]$. (\textit{Difficult!})
\item Does the Fourier sine series of $f$ converges uniformly on $[0, \pi]$?
\begin{proof}
No,it does not converge. Suppose there exists a Fourier sine series that converge to $x$. Then the sine series must be of the form
\[ f(x) = \sum_{n=1}^{\infty} A_n \sin n x\]
Then we compute
\[ f(\pi) = \sum_{n=1}^{\infty} A_n \sin n \pi = 0\]
However, we have $f(\pi) = \pi$. Therefore the sine series does not converge to $x$ on $[0,\pi]$.
\end{proof}
\end{enumerate}
\item Suppose $u = u(x,t)$ solves
\[ \left\{ \begin{array}{ll}
u_t - u_{xx} = f(u) & \mathrm{for} \quad 0 < x <l, t \geq 0 \\
u(0,t) = u(l,t) = 0 & \mathrm{for} \quad t \geq 0\\
\end{array} \right. \]
where $f(s) = -F'(s)$ is a given function. Show that
\[ \frac{d}{dx} \int_0^l \frac 12 u_x^2 (x,t) + F(u(x,t)) \, dx \leq 0.\]
\item Let $u = u(x,t)$ solves
\[ \left\{ \begin{array}{ll}
u_{xx} + u_{yy} = -\lambda u & \mathrm{for} \quad x^2 + y^2 <1 \\
u = 0 & \mathrm{for} \quad x^2 + y^2 = 1 \\
\end{array} \right. \]
where $\lambda > 0$.
\begin{enumerate}
\item State the polar form of the $2$-dimensional Laplace operator.
\item If $u = R(r)\Theta(\theta)$, which ODEs do $R$ and $\Theta$ solve?
\item Solve the ODE for $\Theta$. (\textit{You don't have to solve the more difficult ODE for $R$; its solutions are called Bessel functions.})
\end{enumerate}
\item Let $u = u(x,t)$ solve $u_t + \nabla \cdot \mathbf F = 0$ where $\mathbf F = \mathbf F(\mathbf x) = (F_{1}, F_2, \ldots, F_{d})$ is a given vector field and let $D$ be an open set. Find a formula for
\[ \frac{d}{dt} \int_D u(\mathbf x,t) \, d\mathbf x\]
which only depends on the values of $\mathbf F$ on $\partial D$.
\item Consider the linear wave equation
\[ \left\{ \begin{array}{ll}
u_{tt} - c^2 u_{xx} = f(x,t) & \mathrm{for} \quad - \infty < x < \infty, -\infty < t < \infty \\
u(x,0) = \phi(x) & \mathrm{for} \quad - \infty < x < \infty \\
u_t(x,0) = \psi(x) & \mathrm{for} - \infty < x < \infty
\end{array} \right. \]
\begin{enumerate}
\item Sketch the domain of dependence of the point $(x,t) = (1,1)$ in the $(x,t)$-plane in case $c = 1$. Do the same in the case $c = 2$.
\item State the general solution formula for this problem in terms of the data $f, \phi, \psi$.
\item If $\phi = \psi = 0$ and $f \geq 0$, show that $u \geq 0$.
\item If $\psi = 0$ and $f = 0$, show that $\max_{x,t}|u| \leq \max_x |\phi|$.
\end{enumerate}
\end{enumerate}
\newpage
\[ \phi \quad \varphi \quad \psi \quad \epsilon \quad \varepsilon\]
\end{document}