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\begin{document}


\begin{center}
First Order Equations
\end{center}

\begin{equation*}
(\ast )\qquad a\frac{\partial u}{\partial x}+b\frac{\partial u}{\partial y}=c
\end{equation*}

\bigskip

\textit{Cauchy Data: } Given curve $\Gamma $ in (x,y) and given $u$ on $%
\Gamma $

\bigskip

Introduce new parametric coordinates:

\begin{itemize}
\item $s$ goes out from curve $\Gamma $. So the initial curve is $s=0$

\item $t$ runs along the curve $\Gamma$.
\end{itemize}

$s$ and $t$ need to be independent. For convenience we will usually choose

them to be orthogonal.

\bigskip \noindent We can rewrite $(\ast )$ as a directional derivative in
the direction $s$ away from $\Gamma $ . We then get 
\begin{equation*}
\frac{du}{ds}=\frac{\partial u}{\partial x}\frac{dx}{ds}+ \frac{\partial u} {%
\partial y}\frac{dy}{ds}
\end{equation*}
Similarly differentiating the initial condition along $\Gamma$ (denoted by $%
\prime$) we get

\begin{equation*}
(\ast \ast )\qquad u_0^{\prime}= \frac{\partial u}{\partial x}x_0^{\prime}+ 
\frac{\partial u}{\partial y} y_{0}^{\prime}
\end{equation*}

Combining (*) and (**) we have two equations for $\frac{\partial u}{\partial
x}$ and $\frac{\partial u}{\partial y}$ on $\Gamma$. A solution exists and

is unique if the determinant is nonzero, i.e.

\begin{center}
\emph{Non-characteristic condition:}
\end{center}

\begin{equation*}
\begin{vmatrix}
a & b \\ 
x_{0}^{\prime } & y_{0}^{\prime }%
\end{vmatrix}%
=ay_{0}^{\prime }-bx_{0}^{\prime }\neq 0\qquad \text{all \ \ }s_{1}\leq
s\leq s_{2}
\end{equation*}

\bigskip

\begin{itemize}
\item We therefore solve (*) and (**) with the following procedure

\item paramertize initial curve in terms of $t$.

\item We write the PDE and intial conditions as the characteristic equations

\begin{eqnarray*}
\frac{\partial x}{\partial s} &=&a \\
\frac{\partial y}{\partial s} &=&b \\
\frac{\partial u}{\partial s} &=&c
\end{eqnarray*}

and solve for $x=x(s,t),\quad y=y(s,t), \quad u=u(s,t)$.

\item invert to find $s=s(x,y),\quad t=t(x,y)$ and so $u=u(x,y)$.
\end{itemize}

\begin{center}
example - characteristic IC
\end{center}

\begin{equation*}
\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}=u\qquad u=1\quad 
\text{on }y=x
\end{equation*}

We parameterize the initial conditions by $x=t$, $y=t$, $u=1$. The

characteristic equations become

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

The solution for the initial data is then $x=s+t$ \ , $y=s+t$, Hence, we

cannot invert to find (x,y). Thus, there is NO solution.

Checking directly the characteristic condition

\begin{equation*}
ay_{0}^{\prime }-bx_{0}^{\prime }=1\cdot 1-1\cdot 1=0
\end{equation*}

So the initial curve is characteristic

\bigskip 

\begin{center}
example
\end{center}

\begin{equation*}
\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=0
\end{equation*}

The characteristic curve is $\frac{dy}{dx}=\frac{y}{1}$.

So $y=Ce^{x}$

Hence

$\frac{d}{dx}\left[ u(x,Ce^{x})\right] =\frac{\partial u}{\partial x}+Ce^{x}%
\frac{\partial u}{\partial y}=\frac{\partial u}{\partial x}+y\frac{\partial u%
}{\partial y}$

So%
\begin{equation*}
u(x,Ce^{x})=u(0,Ce^{0})=u(0,C)
\end{equation*}

Let $y=Ce^{x}$ \ \ \ \ \ \ \ \ $C=ye^{-x}$

\begin{eqnarray*}
u(x,y) &=&u(0,e^{-x}y)=f(e^{-x}y) \\
u(0,y) &=&f(y)
\end{eqnarray*}

If $u(0,y)=y^{3}$ initially then

\begin{equation*}
u(x,y)=(e^{-x}y)^{3}=y^{3}e^{-3x}
\end{equation*}

\bigskip 

\begin{center}
example
\end{center}

\begin{eqnarray*}
\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y} &=&1\qquad
-\infty <y<\infty \quad x>0 \\
u(0,y) &=&y^{2}
\end{eqnarray*}

characteristic equations

\begin{eqnarray*}
\frac{\partial x}{\partial s} &=&1 \\
\frac{\partial y}{\partial s} &=&y \\
\frac{\partial u}{\partial s} &=&1
\end{eqnarray*}

So $\frac{dy}{dx}=\frac{\frac{\partial y}{\partial s}}{\frac{\partial x}{%
\partial s}}=y$. \ This has a solution \ $x=s$ and $y=te^{s}$. Reversing we
have

\begin{eqnarray*}
s &=&x \\
t &=&ye^{-s}=ye^{-x}
\end{eqnarray*}

The initial condition is%
\begin{equation*}
x=0\quad y=t\quad u=y^{2}=t^{2}
\end{equation*}

So the characteristic is $y=te^{s}=y_{0}e^{x}$. \ So $y_{0}=ye^{-x}$.

\bigskip 

On the characteristic $\frac{\partial u}{\partial s}=1$ and so $u=s+c(t)$ .

So%
\begin{equation*}
c=u-s=u(0,y)-0=y_{0}^{2}
\end{equation*}

In conclusion 
\begin{equation*}
u(x,y)=s+c=x+y_{0}^{2}=x+(ye^{-x})^{2}=x+y^{2}e^{-2x}
\end{equation*}

\end{document}
