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\begin{document}

\begin{frame}{The method of characteristics}

Recall that the first order linear wave equation
$$
u_t + c u_x = 0, \quad u(x,0) = f(x)
$$
is constant on lines $x - ct = x_0$.  
 
\vspace{.1 in}
\pause
To find $u(x,t)$, go backward along these lines until $t=0$, and
use the initial condition:
$$
u(x,t) = u(x-ct,0) = f(x-ct)
$$

This idea can be extended to many other transport-like equations.

\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


\begin{frame}{Homogeneous transport equations}

Consider
\begin{equation*}
   \label{one}
u_t + c(x,t) u_x = 0, \quad u(x,0) = f(x), \quad -\infty< x< \infty.
\end{equation*}


\vspace{.1 in}
\pause
Let $X(T)$ be some curve in the $(x,t)$ plane.  How does $u$ evolve along
this curve?
$$
\frac{d}{dT} u(X(T),T) = X'(T) u_x( X(T),T) + u_t( X(T), T).
$$ 


\vspace{.1 in}
\pause
If we choose $X'(T) = c(X(T),T)$, then
$$
\frac{d}{dT} u(X(T),T) = c(X(T),T) u_x(X(T),T) + u_t(X(T),T) = 0
$$ 
i.e. $u$ is constant along any such curve.

\vspace{.1 in}
\pause
The curves $X(T)$ that solve the ODE
\begin{equation*}
  \label{char}
X'(T) = c(X,T), \quad X(t) = x,
\end{equation*}
are {\it characteristics} which terminate at $(x,t)$.

\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


\begin{frame}{Example 1.} 

 Solve $u_t + x u_x =0$ with initial condition $u(x,0) = \cos(x)$.

\vspace{.1 in}
\pause
A characteristic curve ending at $(x,t)$ will solve 
$$ X'(T) = X(T), \quad X(t) = x, $$
whose solution is $X(T) = x \exp(T - t)$.

\vspace{.1 in}
\pause
Since $u$ is constant along the characteristic,
$$
u(x,t) = u( X(0), 0) = \cos( x e^{-t}).
$$



\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{Example 2.}  Solve 
\begin{equation*}
   \label{ex2}
y u_x = x u_y, \quad \text{$u(0,y) = 2 y^2$ for $y>0$}
\end{equation*}

\pause
Think of $x$ as time variable and write equation as
$$
u_x - \frac{x}{y} u_y = 0.
$$
Characteristic curves $Y(X)$ solve separable equation
$$
Y'(X) = -\frac{X}{Y},  Y(x) = y.
$$
\pause
Implicit solution is $X^2 + Y^2 = x^2 + y^2$ (i.e. circles)  \pause

\vspace{.1 in}
\pause
Since the solution is constant along curves, setting
$X = 0$ gives
$$
u(x,y) = u(X,Y) = 2 Y^2 = 2(x^2 + y^2).
$$

\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{Inhomogeneous transport equations}

Consider
\begin{equation*}
   \label{two}
u_t + c(x,t) u_x = g(u,x,t), \quad u(x,0) = f(x), \quad -\infty< x< \infty.
\end{equation*}

\pause
$u$ is not constant along characteristics; evolves as
\begin{equation*}
  \label{uode}
   \frac{d}{dt} u(X(t),t) = g(u ,X(T),T).
\end{equation*}

\pause
Let $U(T) = u(X(T),T)$ be solution $u$ restricted to single characteristic,
$$
U'(T) = g(U ,X(T),T), \quad U(0) = u(X(0),0) = f(X(0)).
$$
To find $u(x,t)$, go {\it backwards} along the characteristic 
until $T=0$, and solve this ODE going {\it forwards} to $T=t$.


\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{The general method of characteristics}

Algorithm:
\begin{enumerate}
\item Find the characteristic terminating at $(x,t)$  by solving
$X'(T) = c(X,T)$ subject to $X(t) = x$.  \pause
\vspace{.2 in}
\item
Find the solution along a characteristic by solving
$U'(T) = g(U,X(T),T)$ subject to $U(0) = U(X(0),0)$.  \pause
\vspace{.2 in}
\item Find the solution at the endpoint of the characteristic
by setting $u(x,t) = U(t)$.
\end{enumerate}


\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{Example 1.}
$$
u_t + (x+t) u_x  = t, \quad u(x,0) = f(x).
$$

\pause
Characteristic curves solve the ODE
$$
X'(T) = X + T, \quad X(t) = x.
$$
\pause
Particular solution is $X_p = - T - 1$; general solution is
$X(T) = C e^{T} - T - 1$.  
\pause
Using the condition $X(t) = x$
$$
X (T) = e^{T-t}(x+t+1) - T - 1.
$$

\pause
Determine how $u$ evolves on characteristic:
$$
U'(T) = T, \quad U(0) = f(X(0)) = f(e^{-t}(x+t+1) - 1).
$$

\pause
Solution:
$$
U(T) =  f(e^{-t}(x+t+1) - 1) + \frac12 T^2.
$$

\pause
Solution at $(x,t)$ is 
$$
u(x,t) = U(t) =  f(e^{-t}(x+t+1) - 1) + \frac12 t^2.
$$

\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{Example 2.}
$$
u_t + 3 u_x = -u^2, \quad u(x,0) = f(x). 
$$
\pause
Characteristics solve $X'(T) = 3$ with $X(t) = x$, so that $X = 3(T-t) +x$.

\vspace{.1 in}
\pause
Solution on characteristic evolves as $U'(T) = -U^2(T)$ with $U(0) = f(X(0)) = f(-3 t + x)$.

\vspace{.1 in}
\pause
Separating $dU/U^2 = -dT$ leads to 
$$
U(T) = \frac{1}{T + 1/f(x-3t)}.
$$
Solution is obtained by setting 
$$
u(x,t) = U(t) =  1/[t + 1/f(x-3t) ].
$$

\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{Example 3.}

Consider water flow on landscape of elevation $h(x)$.  

\vspace{.1 in}
Simple model: flow is downhill with magnitude $h'(x)$, which leads to
flux $J = -h'(x) u$ and conservation equation
$$
u_t + ( -h'(x) u)_x = 0.
$$

\vspace{.1 in}
\pause
Put in form of transport equation:
\begin{equation*}
  \label{water}
u_t - h'(x) u_x = h''(x) u.
\end{equation*}


\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{Example 3,cont.}
\begin{small}

Valley described by $h(x) = x^2$, and initial depth is
$$
u(x,0) = \begin{cases} 1 & |x| \le 1 \\ 0 & |x| >1 \end{cases}
$$
so that equation is now
$$
u_t - 2 x u_x = 2 u.
$$
\pause
Characteristics solve $X'(T) = - 2 X$
with terminal condition $X(t) = x$.  \pause Solution:
$$
X(T) = x e^{2(t- T)}.
$$
On characteristics, find $u$ evolves according to $U' = 2U$ and 
$$
U(0) = \begin{cases} 1 & |X(0)| \le 1 \\ 0 & |X(0)| >1 \end{cases}
$$

\pause
Therefore $U(T)  = e^{2 T}$ if $|X(0)| = |x e^{2t}| < 1$, or zero otherwise. \\ \pause
When $t=T$, 
$$
u(x,t) = U(t) = \begin{cases} e^{2t} & |x| \le e^{-2t} \\ 0 &    |x| > e^{-2t}    \end{cases}
$$
\end{small}

\end{frame}

\end{document}









