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\begin{center}
semilinear equation
\end{center}

\bigskip 
\begin{equation*}
\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}=u^{2}
\end{equation*}

Initial condition: \ \ 
\begin{equation*}
u(0,y)=1
\end{equation*}

initial curve:%
\begin{equation*}
x=0,\quad y=t,\quad u=1
\end{equation*}

The characteristic equations are:

\begin{align*}
\frac{\partial x}{\partial s}& =1 \\
\frac{\partial y}{\partial s}& =1 \\
\frac{\partial u}{\partial s}& =u^{2}
\end{align*}

So%
\begin{align*}
x &= s+x_{0}=s \\
y &= s+y_{0}=s+t
\end{align*}

Also%
\begin{align*}
\frac{du}{u^{2}} &= ds \\
-\frac{1}{u} &= s+c \\
u &= -\frac{1}{s+c}=\frac{-1}{s+\frac{1}{u_{0}}}=\frac{u_{0}}{1-su_{0}}=%
\frac{1}{1-s}
\end{align*}

Inverting 
\begin{align*}
s &= x \\
t &= y-x
\end{align*}

So

\begin{equation*}
u=\frac{1}{1-x}
\end{equation*}

Note at $x=1$ we have $u=\infty$.

\bigskip

Homework: consider same equation with IC \ \ \ $u=e^{-y^{2}}$

\newpage

\begin{center}
quasilinear equation
\end{center}

\begin{equation*}
\frac{\partial u}{\partial t}+u\frac{\partial u}{\partial x}=2t
\end{equation*}

Initial condition:
\begin{equation*}
u(x,0)=x=x_{0}
\end{equation*}

The characteristic equations are:

\begin{align*}
\frac{\partial t}{\partial s}& =1 \\
\frac{\partial x}{\partial s}& =u\text{\qquad  So } \frac{dx}{dt}=u \\
\frac{\partial u}{\partial s}& =2t
\end{align*}

Note,  that the characteristic curves now depend on $u$ .

However, the first equation integrates to $t=s$ (remember $t$ is time). So

$\frac{\partial u}{\partial s}= 2t$ \ becomes 
$\frac{\partial u}{\partial t} = 2t$ 
\begin{equation*}
u(x,t)=t^{2}+C\qquad C=u-t^{2}=u(x,0)=x_{0}
\end{equation*}

The slope of the characteristics is now given by%
\begin{equation*}
\frac{dx}{dt}=u=x_{0}+t^{2}
\end{equation*}

So the characteristic curve is given by%
\begin{align*}
x &= \frac{1}{3}t^{3}+x_{0}(t+1) \\
\text{or\quad\ }x_{0} &= \frac{x-\frac{1}{3}t^{3}}{(t+1)}
\end{align*}

On the characteristic 
\begin{equation*}
C=x_{0}=\frac{3x-t^{2}}{3\left( t+1\right) }
\end{equation*}

So%
\begin{equation*}
u(x,t)=t^{2}+C=t^{2}+\frac{3x-t^{2}}{3\left( t+1\right) }
\end{equation*}

So in this case the solution is well defined for all $t>0$

\newpage

\begin{center}
quasilinear equation
\end{center}

\bigskip

\begin{equation*}
\frac{\partial u}{\partial t}+a(u)\frac{\partial u}{\partial x}=0
\end{equation*}

\noindent Represents fluid flow and traffic flow. 
\medskip 
\newline
\noindent So $\frac{du}{dt}=0$ \ (i.e. $u$ is constant) along the
characteristics $\frac{dx}{dt}=a(u)$. \medskip \newline
\noindent Since $u$ is constant along the characteristic so is $a(u)$.
\medskip 
\newline
\noindent Hence $\frac{dx}{dt}$ is constant and so the characteristics 
are straight lines.

So%
\begin{equation*}
u(x,t)=f(x-a(u)t)
\end{equation*}

\noindent This an implicit solution for $u$. Differentiating we get%
\begin{equation*}
u_t = \frac{-a}{1+ f^{\prime} a_u t} f^{\prime} \quad 
u_x = \frac{1} {1+ f^{\prime} a_u t} f^{\prime}
\end{equation*}

Assume $a_{u}>0$ \quad In particular $a_{u}\neq 0$ .
\noindent The slope is 
$\frac{dt}{dx}=\frac{1}{a(u(x_{0},t_{0})}=\frac{1}{a(f(x_{0}))}$.

\bigskip

case I: \ \quad $f^{\prime} > 0$. Then u at $t=0$ increases
and the slope $\frac{1}{u}$ decreases. So the solution and its 
derivatives are bounded.

case II: \quad $f^{\prime} < 0$. 
Then u at $t=0$ decreases
and the slope $\frac{1}{u}$ increases. So the solution and its 
derivatives are unbounded. $u_{x}$ and $u_{t}$ approach
infinity as $t\rightarrow \frac{-1}{u_0^{\prime} a_u}= 
\frac{1}{ \vert u_{0}^{\prime} \vert a_u}$.

\bigskip

\noindent So we have two cases:

\begin{itemize}
\item expansion wave \qquad $a(f(z)) \!\geq \! a(f(w))$ 
\ whenever $z \!\leq \!w$. Solution and derivatives are bounded.

\item compression wave = shock wave. Solution and derivatives may
become unbounded.
\end{itemize}

\bigskip 
\noindent Physically this means that later air moves faster than
earlier air. 
\medskip 
\newline
\noindent Hence, this \emph{seems} to imply that there is no 
solution beyond some $t_{\ast }$. 
\medskip 
\newline
\noindent However, physically the fluid continues to exist 
beyond the singularity! 
\medskip 
\newline
\noindent Hence, we need to define a generalized solution 
that allows solutions for all time.

\end{document}
