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\title{Midterm 1}
\date{}


\theoremstyle{definition} 
\newtheorem{definition}{Definition}[section]
\newtheorem{conjecture}{Conjecture}[section]
\newtheorem{example}{Example}[section]
\newtheorem{lemma}{Lemma}[section]
\newtheorem{remark}{Remark}[section]
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\begin{document}
\maketitle
\begin{enumerate}
\item Solve the first order PDE
\begin{equation*}
\begin{cases}
-2u_x+u_y-yu=0 \\
u(x,0) = x^2.
\end{cases}
\end{equation*}
\item Attempt to find a particular solution of 
\begin{equation*}
3yu_x-2xu_y=0.
\end{equation*}  Determine if there is a unique solution, no solution, or infinitely many solutions for auxiliary conditions
\begin{enumerate}
\item $u(x,y) = x^2$ on the line $y=x$.
\item $u(x,y) = 1-x^2$ on the line $y=-x$.
\item $u(x,y) = 2x$ on the ellipse $2x^2+3y^2=4$.
\end{enumerate}

\item Show that the general solution of the wave equation $u_{tt}-c^2u_{xx}=0$ is given by 
\begin{equation*}
u(x,t) = f(x-ct) + g(x+ct)
\end{equation*}
for sufficiently differentiable functions $f$ and $g$ by considering the change of variables
\begin{align*}
s &= x-ct\\
r &= x+ct.
\end{align*}
\item Solve the initial value problem 
\begin{equation*}
\begin{cases}
u_{tt}-u_{xx}=0 & \text{ on } -\infty < x < \infty, t> 0\\
u(x,0) = 0\\
u_t(x,0) = \psi(x).
\end{cases}
\end{equation*}
where
\begin{equation*}
\psi(x)= \begin{cases}
1 & \text{ on } |x|\leq 1\\
0 & \text{ on } |x|>1.
\end{cases}
\end{equation*}
You may leave your solution in terms of $\psi$.  Using this solution, answer the following
\begin{enumerate}
\item Will the wave ever return to its original state at $x=0$?
\item What happens at each point $x$ when $t\to \infty$?
\item For some positive time $t>0$, is there a location $x$ such that $u(x,t) = 0$?
\end{enumerate}
\end{enumerate}

\newpage

\color{red}
\section{Solutions}

\begin{enumerate}
\item First we compute the characteristic curves.
\begin{align*}
\frac{d x}{d y} = -2.
\end{align*}
Solving for this, we have
\begin{align*}
x = -2y+t
\end{align*}
Changing variables so that
\begin{align*}
s&= y\\
t&= x+2y
\end{align*}
transforms the PDE to
\begin{align*}
u_s-su=0.
\end{align*}
This can be solved as a separable ODE so that
\begin{align*}
\frac{du}{u} = sds
\end{align*}
so
\begin{align*}
\ln|u| = \frac{1}{2}s^2+f(t)
\end{align*}
where $f$ is some function that depends on the other variable $t$.  Exponentiating both sides yields
\begin{align*}
u(s,t) = C(t)e^{\frac{s^2}{2}}.
\end{align*}
Plugging back $x$ and $y$, we get
\begin{align*}
u(x,y) = C(x+2y)e^{\frac{y^2}{2}}.
\end{align*}
Plugging in the auxiliary condition, we have
\begin{align*}
x^2=u(x,0) = C(x),
\end{align*}
which tells us what the function $C(t)$ is hence the particular solution is given by
\begin{align*}
u(x,y) = (x+2y)^2e^{\frac{y^2}{2}}.
\end{align*}

\item First we find the general solution by the method of characteristic.  Dividing by $3y$, we get $u_x-\frac{2x}{3y}u_y = 0$.  Then we want to solve the ODE
\begin{align*}
\frac{dy}{dx} = -\frac{2x}{3y}.
\end{align*}
This gives us
\begin{equation*}
\frac{3}{2}y^2= -x^2+t.
\end{equation*}
For convenience, we multiply through by 2 and use the change of variables
\begin{align*}
s&=x\\
t&=2x^2+3y^2.
\end{align*}
The change of variables transforms the PDE to 
\begin{align*}
u_s = 0,
\end{align*}
hence the solution is given by
\begin{align*}
u(s,t) = f(t)
\end{align*}
or
\begin{align*}
u(x,y) = f(2x^2+3y^2).
\end{align*}
Now we check each condition.
\begin{enumerate}
\item $x^2= u(x,x) = f(5x^2)$, hence $f(s)=\frac{s}{5}$ would satisfy the initial condition so $u(x,y) = \frac{2x^2+3y^2}{5}$.
\item $1-x^2=  u(x,-x) = f(5x^2)$ so $f(s) = 1-\frac{s}{5}$.  Therefore the particular solution is $u(x,y) = 1-\frac{2x^2+3y^2}{5}$.
\item $2x = u(x,y) = f(4)$, since the right hand side is a constant and the left a variable, there does not exist an $f(s)$ that satisfies the equation, hence no solution. 
\end{enumerate}
\item By applying Chain rule, the change of variables transforms the PDE to
\begin{equation*}
u_{sr} = 0
\end{equation*}
This means that
\begin{equation*}
u(s,r) = f(s) + g(r)
\end{equation*}
for some functions $f$ and $g$.  Plugging back the variables, we have
\begin{equation*}
f(x,y) = f(x-ct)+g(x+ct).
\end{equation*}
\item By d'Alembert's formula, the solution is given by
\begin{equation*}
u(x,y) = \frac{1}{2}\int_{x-t}^{x+t}\psi(s)ds.
\end{equation*}
\begin{enumerate}
\item No, see part $b$ for details.
\item Fix an arbitrary $x$.  Then taking the limit as $t\to \infty$, we have
\begin{equation*}
\lim_{t\to \infty}u(x,t) = \frac{1}{2}\int_{-\infty}^\infty \psi(s)ds = \frac{1}{2}\int_{-1}^1\psi(s)ds = 1.
\end{equation*}
So each point will converge to the height $1$.  Since the effect holds for any $x$, in particular, at $x=0$, the height will converge to 1 and hence will not return to its initial height, $u(0,0) = 0$.
\item Simply consider the point $x=2$ and $t=1$. Then 
\begin{equation*}
u(2,1) = \int_1^3\psi(s)ds = 0.
\end{equation*}
In general, choosing a point $x>t+1$ would yield 0 height.  This comes from the fact that the solution to the wave equation experiences finite propagation.  
\end{enumerate}
\end{enumerate}
\end{document}


