% ch01slidesa.tex

\begin{slide}
\begin{center}\large\bf
Special Topics from Asmar's Textbook, Chapter 1
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\begin{itemize}
\item {Method of Characteristics}

\end{itemize}

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\begin{slide}
\header{Method of Characteristics}

\bigskip

\BF{Definition}.  A first order partial differential equation 
\begin{equation}\label{first-order-pde}
v_1(x,y)\frac{\partial u(x,y)}{\partial x} + v_2(x,y)\frac{\partial u(x,y)}{\partial y} =0\end{equation}
has \DF{characteristic curves} defined by the implicit solution
$$w(x,y)=c$$
of the associated \DF{characteristic differential equation}
$$v_1(x,y)dx + v_2(x,y)dy = 0 .$$

\begin{Theorem}[General Solution]
Let $f(w)$ denote an arbitrary function. Then the general solution of (\ref{first-order-pde}) is
given by
$$u(x,y) = f(w(x,y)).$$
\end{Theorem}
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\begin{slide}
\header{General Solution by the Method of Characteristics}

\bigskip

\BF{Proof}:
Let $f(w)$ denote an arbitrary function. We prove that the general solution of (\ref{first-order-pde}) is
given by $u(x,y) = f(w(x,y))$. First, suppose that $(x_0,y_0)$ is a point of the characteristic curve $w(x,y)=c$
and $y$ is locally determined as a function of $x$, e.g., $v_1(x,y)\ne 0$ and $y=y(x)$. Then $y(x)$ is differentiable and
$y'=v_2/v_1$. Assume $u(x,y)$ is a solution of (\ref{first-order-pde}), then we compute
$$\begin{array}{lcl} \dd\frac{d}{dx} u(x,y(x)) &=& \dd\frac{\partial u}{\partial x} + y'(x)\frac{\partial u}{\partial y} \\
&=& \dd\frac{1}{v_1(x,y)}\left(v_1(x,y)\frac{\partial u}{\partial x} + v_2(x,y)\frac{\partial u}{\partial y}\right) \\
&=& 0. \end{array}
$$
If the derivative is zero, then $u(x,y(x))$ must be a constant which depends only on $(x_0,y_0)$, or ultimately on the constant $c$ in the equation $w(x_0,y_0)=c$. Therefore, $u(x,y(x))=f(c)$ for some function $f(w)$. Using the implicit solution, then $u(x,y(x))=f(c)=f(w(x,y(x))$ or simply $u(x,y)=f(w(x,y))$. The proof is completed by showing directly that this solution satisfies the partial differential equation.


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