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\noindent
Note from (9.1) and (9.2) that
\[ \alpha \phi_x + \beta \phi_y = \bf{u} \cdot
\nabla \phi = \left. \dfrac{\phi}{s} \right|_C \]
is the directional derivative along integral
curves $C$ of $\bf{u} = (\alpha, \beta)$, called
the \emph{characteristic curves} of the PDE.
Since $\dfrac{\phi}{s} = \alpha \phi_x + \beta
\phi_y = 0$ from (9.4), the function $\phi(x, y)$
will be constant along the curves $C$, i.e. the
Cauchy data $f(t)$ defined on $B$ at $s = 0$ will
be propagated constantly along the curve $C$ to
give solution
\[ \phi(s, t) = \phi(x(s, t), y(s, t)) = f(t)
\tag{9.6} \]
To obtain $\phi(x, y)$ transform coordinates from
$\phi(t, s)$ using $s = s(x, y)$, $t = t(x, y)$
(provided Jacobian $J = x_t y_s - x_s y_t \neq 0$)
to finally obtain
\[ \phi(x, y) = f(t(x, y)) \tag{9.7} \]
Prescription: To solve (9.4) with (9.5)
\begin{enumerate}[(1)]
    \item Find characteristic equation (9.3)
        $\dfrac{x}{s} = \alpha$, $\dfrac{y}{s} =
        \beta$.
    \item Parametrise initial conditions on $B$
        $(x(t), y(t))$ (9.8)
    \item Solve characteristic equation (9.3) to
        find $x(s, t)$ and $y(s, t)$ subject to
        (9.8) at $s = 0$, $x(0, t) = x(t)$, $y(0,
        t) = y(t)$.
    \item Solve (9.4) with (9.1)
        \[ \dfrac{\phi}{s} = \alpha \phi_x + \beta
        \phi_y = 0 \]
        (9.6) $\phi(s, t) = f(t)$ [or $\gamma(s,
        t)$ on RHS].
    \item Invert relations $s = s(x, y)$, $t =
        t(x, y)$.
    \item Change coordinates to obtain (9.7),
        $\phi(x, y)$.
\end{enumerate}

\begin{flashcard}[first-order-characteristics-example]
\begin{example*}
    Solve $e^x \phi_x + \phi_y = 0$ with $\phi(x,
    0) = \cosh x$.
    \begin{enumerate}[(1)]
        \item \cloze{Characteristic equations
            \[ \dfrac{x}{s} = e^x, \qquad \dfrac{y}{s} =
            1 \tag{$*$} \]}%
        \item \cloze{Initial conditions $x(t) = t$, $y(t)
            = 0$ on the $x$ axis ($\dag$).}
        \item \cloze{From ($*$), $\frac{\dd x}{e^x} = \dd
            s$, $-e^{-x} = s + c$, $y = s + d$. At
            $s = 0$, $x = 0$, $-e^{-t} = c$, $y =
            0 = d$.
            \[ -e^{-x} = e^{-t} - s, \qquad y = s \]
            (characteristics).}
        \item \cloze{$\dfrac{\phi}{s} = 0 \implies
            \phi(s, t) = \cosh t$.}
        \item \cloze{$s = y$, $e^{-t} = y + e^{-x}
            \implies t = -\log(y + e^{-x})$.}
        \item \cloze{So
            \[ \phi(x, y) = \cosh[-\log(y +
            e^{-x})] \]}%
    \end{enumerate}
\end{example*}
\end{flashcard}

\subsubsection*{Inhomogeneous 1st order PDE}

Want to solve
\[ \alpha(x, y) \phi_x + \beta(x, y) \phi_y =
\gamma(x, y) \tag{9.9} \]
with Cauchy data $\phi(x(t), y(t)) = f(t)$ on
curve $B$. The characteristic curves $C$ are
identical to homogeneous case (9.4) but now (9.1)
implies
\[ \left. \dfrac{\phi}{s} \right|_C = \bf{u} \cdot
\nabla \phi = \gamma(x, y) \tag{9.10} \]
with $\phi = f(t)$ at $s = 0$ on $B$, i.e. no
longer propagating constantly and we must solve an
ODE (9.10). So upgrade point 4 in prescription to
integrate $\phi(s, t)$ before reverting to
$\phi(x, y)$.

\begin{example*}
    Solve $\phi_x + 2\phi_y = ye^x$ with $\phi =
    \sin x$ on $y = x$.
    \begin{enumerate}[(1)]
        \item Characteristic equation
             \[ \dfrac{x}{s} = 1, \qquad \dfrac{y}{s} =
             2 \tag{$*$} \]
        \item So on $y = x$, take $(x(t), y(t)) =
            (t, t)$ ($\dag$).
        \item From ($*$), $x = s + c$, $y = 2s +
            d$. So because of ($\dag$), $s = 0$,
            $x = t = c$, $y = t = d$.
            \[ x = s + t, \qquad y = 2s + t \]
        \item Solve $\dfrac{\phi}{s} = \gamma =
            ye^x = (2s + t) e^{s + t}$ with $\phi
            = \sin t$ at $s = 0$. Note
            $\dfrac{}{s} (2s e^s) = 2e^s + 2se^s$
            so
            \[ \phi(s, t) = (2s - 2 + t) e^{s + t}
            + \text{const} \]
            But at using $s = 0$ condition we have
            $\phi(0, t) = \sin t = (t - 2)e^t +
            \text{const}$ so
            \[ \phi(s, t) = (2s - 2 + t)e^{s + t}
            + \sin t + (2 - t)e^t \]
        \item Invert $s = y - x$, $t = 2x - y$.
        \item So
            \[ \phi(x, y) = (y - 2) e^x + (y - 2x +
            2)e^{2x - y} + \sin(2x - y) \]
    \end{enumerate}
\end{example*}

\subsection{Second-order PDE classification}

In two dimensions, the general 2nd order linear
PDE is
\[ \mathcal{L} \equiv a(x, y)
\pfrac[2]{\phi}{x} + 2b(x, y) \frac{\partial^2
\phi}{\partial x \partial y} + c(x, y)
\pfrac[2]{\phi}{y} + d(x, y) \pfrac{\phi}{x} +
e(x, y) \pfrac{\phi}{y} + f(x, y) \phi(x, y) = 0 \]
\begin{hiddenflashcard}[second-order-PDE-form]
\prompt{Form of linear second order PDE for method
of characteristics?}
\[ \mathcal{L} \equiv \cloze{a\phi_{xx}
+ 2b \phi_{xy} + c \phi_{yy} + d\phi_x
+ c\phi_y + f\phi = 0} \]
\end{hiddenflashcard}
The \emph{principal part} is given by
\begin{align*}
    \sigma_p(x, y, k_x, k_y)
    &\equiv K^\top AK \\
    &= (k_x ~ k_y)
    \begin{pmatrix}
        a(x, y) & b(x, y) \\
        b(x, y) & c(x, y)
    \end{pmatrix}
    \begin{pmatrix}
        k_x \\
        k_y
    \end{pmatrix}
\end{align*}
The PDE is classified by the eigenvalues of $A$:
\begin{itemize}
    \item $b^2 - ac < 0$ \emph{elliptic}
        ($\lambda_1, \lambda_2$ same sign)
    \item $b^2 - ac > 0$ \emph{hyperbolic}
        ($\lambda_1, \lambda_2$ opposite sign)
    \item $b^2 - ac$ \emph{parabolic} ($\lambda_1$
        or $\lambda_2 = 0$)
\end{itemize}
Exercise: show this from $\det(A - \lambda I)$,
i.e. $\lambda_\pm = \half (\Trace \pm
\sqrt{\Trace^2 - 4 \det})$.

\subsubsection*{Examples}

\begin{itemize}
    \item Wave equation (3.4)
        \[ \frac{1}{c^2} \pfrac[2]{\phi}{t} =
        \pfrac[2]{\phi}{x} \]
        $a = \frac{1}{c^2}$, $b = 0$, $c = -1$ is
        hyperbolic.
    \item Heat equation (4.3) $a = 0$, $b = 0$, $c
        = -D$ is parabolic.
    \item Laplace equation (5.1) $a = 1$, $b = 0$,
        $c = 1$ is elliptic.
\end{itemize}

\subsubsection*{Characteristic curves}

A curve defined by $f(x, y) = 0$ will be a
characteristic curve if
\[ (f_x ~ f_y)
\begin{pmatrix}
    a & b \\
    b & c
\end{pmatrix}
\begin{pmatrix}
    f_x \\
    f_y
\end{pmatrix}
= 0 \tag{9.12} \]
(generalisation 1st order $\nabla f \cdot \bf{u} =
0$, $\bf{u} = (\alpha, \beta)$). The curve can be
written as $y = y(x)$ where
\[ \pfrac{f}{x} + \pfrac{f}{y} \dfrac{y}{x} = 0
\implies \frac{f_x}{f_y} = -\dfrac{y}{x}
\tag{9.13} \]
Substituting into (9.12) we obtain a quadratic
with solution
\begin{flashcard}[second-order-characteristic-curve]
\prompt{How to find characteristic curves for
linear second order PDE?}
\[ \cloze{\dfrac{y}{x} = \frac{b \pm \sqrt{b^2 -
ac}}{a}} \tag{9.14} \]
\end{flashcard}
(exercise).
