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\begin{itemize}
    \item Hyperbolic if $b^2 - ac > 0$, then
        \emph{2 solutions}
    \item Parabolic if $b^2 - ac = 0$, then
        \emph{1 solution}
    \item Elliptic if $b^2 - ac < 0$, \emph{no
        real solutions}
\end{itemize}
Transforming to characteristic coordinates $(u,
v)$ would set $a = c = 0$ in (9.11) so the PDE
takes \emph{canonical form}
\[ \frac{\partial^2 \phi}{\partial u \partial v} +
\cdots = 0 \tag{9.15} \]
where the dots would be lower-order terms $\phi_u,
\phi_v, \phi$ (Refer to section 9.4 in R Josza
lecture notes).

\begin{example*}
    Consider
    \[ -y \phi_{xx} + \phi_{yy} = 0 \tag{$*$} \]
    With $a = -y$, $b = 0$, $c = 1$, $b^2 - ac =
    y$. So hyperbolic for $y > 0$ (elliptic for $y
    < 0$, parabolic for $y = 0$). Find
    characteristics for $y > 0$ satisfying (9.14)
    \[ \dfrac{y}{x} = \frac{b \pm \sqrt{b^2 -
    ac}}{a} \pm \frac{1}{\sqrt{y}} \implies
    \sqrt{y} \dd y = \pm \dd x \]
    \[ \implies \frac{2}{3} y^{3/2} \pm x = c_\pm \]
    so characteristic curves are
    \[ u = \frac{2}{3} y^{3/2} + x, \qquad v =
    \frac{2}{3} y^{3/2} - x \]
    Derivatives are $u_x = 1$, $u_y = y^{1/2}$,
    $v_x = -1$, $v_y = y^{1/2}$. Hence
    \[ \phi_x = \phi_u u_x + \phi_v v_x = \phi_u -
    \phi_v \]
    \[ \phi_y = y^{1/2} (\phi_u + \phi_v) \]
    \[ \phi_{xx} = \phi_{uu} - 2\phi_{uv} +
    \phi_{vv} \]
    \[ \phi_{yy} = y(\phi_{uu} + 2\phi_{uv} +
    \phi_{vv}) + \frac{1}{2y^{1/2}}(\phi_u +
    \phi_v) \]
    From ($*$)
    \[ -y\phi_{xx} + \phi_{yy} = y(4\phi_{uv} +
    \frac{1}{2y^{3/2}}(\phi_u + \phi_v) = 0 \]
    Now using $u + v = \frac{4}{3} y^{3/2}$ and $y
    > 0$, the canonical form is
    \[ \phi_{uv} + \frac{1}{6(u + v)}(\phi_y +
    \phi_v) = 0 \]
\end{example*}

\subsection{General solution for Wave Equation
(D'Alembert)}

Solve (3.4)
\[ \frac{1}{c^2} \pfrac[2]{\phi}{t} -
\pfrac[2]{\phi}{x} = 0 \]
with initial conditions
\[ \phi(x, 0) = f(x), \qquad
\phi_t(x, 0) = g(x) \tag{9.16} \]
With $a = \frac{1}{c^2}$, $b = 0$, $c = -1$ the
characteristic equation
\[ \dfrac{x}{t} = \frac{-0 \pm \sqrt{0 +
\frac{1}{c^2}}}{\frac{1}{c^2}} = \pm c \]
so choose $u = x - ct$ and $v = x + Ct$, which
yields simple canonical form:
\[ \frac{\partial^2 \phi}{\partial u\partial v} =
0 \tag{9.17} \]
Integrate with respect to $u$, $\pfrac{\phi}{v} =
F(v)$ and then with respect to $u$
\[ \phi = G(u) + \int^v F(y) \dd y = G(u) + H(v) \]
Impose our initial conditions at $t = 0$ when $u =
v = x$,
\[ \phi(x, 0) = G(x) + H(x) = f(x) \tag{$*$} \]
\[ \phi_t(x, 0) = -cG'(x) + cH'(x) = g(x)
\tag{$\dag$} \]
Differentiating ($*$):
\[ G'(x) + H'(x) = f'(x) \tag{$\ddag$} \]
So ($\dag$) and ($\ddag$)
\[ \implies H'(x) = \half (f'(x) +
\frac{1}{c}g(x)) \]
Integrate
\[ H(x) = \half(f(x) - f(0)) + \frac{1}{2c}
\int_0^x g(y) \dd y \]
and from ($*$)
\[ G(x) = \half (f(x) - f(0)) - \frac{1}{2c}
\int_0^x g(y) \dd y \]
Putting together:
\begin{align*}
    \phi(x, t)
    &= G(x - ct) + H(x + ct) \\
    &= \half (f(x - ct) + f(x + ct)) +
    \frac{1}{2c} \int_{x - ct}^{x + ct} g(y) \dd y
\end{align*}

\newpage
\section{Solving PDEs with Green's Functions}

\subsection{Diffusion equation and Fourier
transform}

Recall heat equation (4.3) for a conducting wire
\[ \pfrac{\theta}{t}(x, t) - D
\pfrac[2]{\theta}{t} (x, t) = 0 \tag{10.1} \]
with initial conditions $\theta(x, 0) = h(x)$ with
$\theta \to 0$ as $x \to \pm \infty$. Take the
Fourier Transform with respect to $x$ using (8.13)
\[ \pfrac{}{t} \tilde{\theta}(k, t) = -D k^2
\tilde{\theta}(k, t) \]
Integrate $\tilde{\theta}(k, t) = C e^{-Dk^2t}$
with initial conditions $\tilde{\theta}(k, 0) =
\tilde{h}(k)$, we have
\[ \tilde{\theta}(k, t) = \tilde{h}(k) e^{-Dk^2t} \]
Now invert
\begin{align*}
    \theta(x, t)
    &= \frac{1}{2\pi}
    \int_{-\infty}^\infty \tilde{h}(k)
    \ub{e^{-Dk^2t}}_{\text{Gaussian (8.5)}}
    e^{ikx} \dd k \\
    &= \frac{1}{\sqrt{4\pi Dt}}
    \int_{-\infty}^\infty h(u) \exp \left(
    \frac{-(x - u)^2}{4Dt} \right) \dd u
    &&\text{by convolution theorem (8.17)} \\
    &\equiv \int_{-\infty}^\infty h(u) S_d(x - u,
    t) \dd u \tag{10.2}
\end{align*}
where the \emph{fundamental solution} is
\[ S_d(x, t) = \frac{1}{\sqrt{4\pi Dt}}
e^{-x^2/4Dt} \tag{10.3} \]
(Fourier transform is $\tilde{S}_d(k, t)
=e^{-Dk^2t}$). Also known as diffusion kernel or
source.

\begin{note*}
    With localised initial conditions $\theta(x,
    0) = \theta_0 \delta(x)$ then
    \[ \theta(x, t) = \theta_0 S_d(x, t) =
    \frac{\theta_0}{\sqrt{4\pi Dt}} e^{-\eta^2}
    \tag{10.4} \]
    where $\eta = \frac{x}{2\sqrt{Dt}}$ is the
    similarity parameter. Initial condition $t \ge
    0$ spreads smoothly as a Gaussian.
\end{note*}

\begin{hiddenflashcard}[Green-function-for-heat-eq]
\prompt{Green's function for homogeneous heat equation}
\[ \cloze{S_d(x, t) = \frac{1}{\sqrt{4\pi Dt}}
e^{-\eta^2}} \]
\prompt{\cloze{where $\eta$ is the similarity
parameter. Note that this function is constructed
to satisfy the homogeneous heat equation with
initial conditions $\theta(x, 0) = \delta(x)$.}}
\end{hiddenflashcard}

\begin{example*}[Gaussian pulse]
    Suppose initially
    \[ f(x) = \sqrt{\frac{a}{\pi}} \theta_0
    e^{-ax^2} \]
    \begin{align*}
        \theta(x, t)
        &= \frac{\theta_0\sqrt{a}}{\sqrt{4\pi^2 Dt}}
        \int \exp \left[ -au^2 -
        \frac{(x - u)^2}{4Dt} \right] \dd u \\
        &= \frac{\theta_0 \sqrt{a}}{\sqrt{4\pi^2
        Dt}} \int_{-\infty}^\infty \exp \left[
        \frac{(1 + 4aDt) u^2 - 2xu + x^2}{4Dt}
        \right] \dd u \\
        &= \frac{\theta_0\sqrt{a}}{\sqrt{4\pi^2
        Dt}} \int_{-\infty}^\infty \exp \left[
        -\frac{(1 + 4aDt)}{4Dt} \left( u -
        \frac{x}{1 + 4Dt} \right)^2 \right] \dd u
        \times \exp \left[ \frac{-ax^2}{1 + 4aDt}
        \right] \\
        &= \theta_0 \sqrt{\frac{a}{\pi(1 + 4Dt)}}
        \exp \left[ \frac{-ax^2}{1 + 4aDt} \right]
        \tag{10.5}
    \end{align*}
    Here, width spreads as standard deviation
    $\propto \sqrt{t}$ with area constant (i.e.
    heat energy conserved).
\end{example*}
