% vim: tw=50 % 23/11/2022 10AM \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).