% vim: tw=50
% 21/10/2022 10AM

\noindent
But $\frac{\ddot{T}}{T}$ depends only on $t$, and
$\frac{X''}{X}$ depends only on $x$! \\
So both sides must be equal to a constant, say
$-\lambda$, so
\[ X'' + \lambda X = 0 \tag{3.6} \]
\[ \ddot{T} + \lambda c^2 T = 0 \tag{3.7} \]

\subsection{Boundary conditions and normal modes}

Three possibilities for $\lambda$ (+, 0, -) in
spatial ODE (3.6) but restricted by (3.1)
\begin{enumerate}[(i)]
    \item $\lambda < 0$. Take $\chi^2 = -\lambda$
        then
        \[ X(x) = Ae^{\chi x} + B e^{-\chi x} =
        \tilde{A} \cosh \chi x + \tilde{B} \sinh
        \chi x \]
        but boundary conditions imply $X(0) = X(L)
        = 0 \implies \tilde{A} = \tilde{B} = 0$
        (only trivial solution works).
    \item $\lambda = 0$ then $X(x) = Ax + B$ but
        then by boundary conditions $A = B = 0$.
    \item $\lambda > 0$, then $X(x) = A \cos
        \sqrt{\lambda} x + B\sin \sqrt{\lambda}
        x$. Here, the boundary conditions (3.1)
        imply $A = 0$ and $B \sin\sqrt{\lambda} L
        = 0$, so $\sqrt{\lambda} L = n\pi$. So
        \[ X_n(x) = B_n \sin \frac{n\pi x}{L},
        \quad \lambda_n = \left( \frac{n\pi}{L}
        \right)^2  \tag{3.8} \]
        i.e. eigenfunctions and eigenvalues of the
        system.
\end{enumerate}
These are \emph{normal modes} because spatial
shape in $x$ does not change in time (amplitude
may vary).
\begin{itemize}
    \item Fundamental mode ($n = 1$): $\lambda_1 =
        \frac{\pi^2}{L^2}$. Lowest frequency
        vibration or first harmonic.
        \begin{center}
            \includegraphics[width=0.6\linewidth]
            {images/a2019e9c512211ed.png}
        \end{center}
    \item Second mode ($n = 2$): $\lambda_2 =
        \frac{4\pi^2}{L^2}$ second harmonic or
        overtone.
        \begin{center}
            \includegraphics[width=0.6\linewidth]
            {images/a8a9e812512211ed.png}
        \end{center}
    \item Third mode $n = 3$ etc
        \begin{center}
            \includegraphics[width=0.6\linewidth]
            {images/af4d8d68512211ed.png}
        \end{center}
\end{itemize}

\subsection{Initial conditions and temporal
solution}

Substitute eigenvalues $\lambda_n = \left(
\frac{n\pi}{L} \right)^2$ into time ODE (3.7)
\[ \ddot{T} + \frac{n^2 \pi^2 c^2}{L^2} T = 0 \]
which has solutions
\[ T_n(t) = C_n \cos \frac{n\pi ct}{2} + D_n \sin
\frac{n\pi ct}{L} \tag{3.9} \]
Thus a specific solution to (3.4) satisfying
boundary conditions (3.1) is
\begin{align*}
    y_n(x, t)
    &= T_n(t) X_n(x) \\
    &= \left( C_n \cos \frac{n\pi ct}{L} + D_n \sin
    \frac{n\pi ct}{L} \right) \sin \frac{n\pi x}{L}
\end{align*}
(absorbing $B_n$ into $C_n$ and $D_n$). Exercise:
verify that this is a solution. \myskip
Since the wave equation (3.4) is linear (and
boundary conditions (3.1) are homogeneous) we can
add the solutions together to find \emph{general
string solution}
\begin{flashcard}[general-string-solution]
\prompt{General string solution?}
\[ \boxed{y(x, t) = \cloze{\sum_{n = 1}^\infty \left( C_n \cos
\frac{n\pi ct}{L} + D_n \sin \frac{n\pi c t}{L}
\right) \sin \frac{n\pi x}{L}} \tag{3.10}} \]
\end{flashcard}
By construction (3.10) satisfies boundary
conditions (3.1), so now impose initial conditions
(3.2): \\
For $t = 0$ we have
\[ y(x, 0) = p(x) = \sum_{n = 1}^\infty C_n \sin
\frac{n\pi x}{L} \]
by (3.10) and also by (3.10):
\[ \pfrac{y}{t} (x, 0) = q(x) = \sum_{n =
1}^\infty \frac{n\pi c}{L} D_n \sin \frac{n\pi
x}{L} \]
So the coefficients are those for Fourier sine
series given by (1.12):
\[ C_n = \frac{2}{L} \int_0^L p(x) \sin \frac{n\pi
x}{L} \dd x \]
\[ D_n = \frac{2}{2\pi c} \int_0^L q(x) \sin
\frac{n\pi x}{L} \dd x \tag{3.11}\]
Hence (3.10-11) is the solution to (3.4)
satisfying (3.1-2).

\begin{example*}
    Pluck string at $x = 3$, drawing it back as
    \[ y(x, 0) = p(x) = \begin{cases}
        x(1 - \xi) & 0 \le x \le \xi \\
        \xi (1 - x) & \xi < x \le 1
    \end{cases} \]
    \[ \pfrac{y}{t} (x, 0) = q(x) = 0 \]
    Then with Fourier series (1.8)
    \[ C_n = \frac{2\sin n\pi \xi}{(n\pi)^2},
    \quad D_n = 0 \]
    so we have solution
    \[ y(x, t) = \sum_{n = 1}^\infty
    \frac{2}{(n\pi)^2} \sin n\pi\xi \sin n\pi x
    \cos n\pi ct \]
    Take $\xi = \half$ then $C_{2m} = 0$, $C_{2m -
    1} = \frac{2(-1)^{m + 1}}{((2m - 1)\pi)^2}$.
    For a guitar, $\frac{1}{4} \le \xi \le
    \frac{1}{3}$, for a violin, $\xi \approx
    \frac{1}{7}$.
\end{example*}

\subsubsection*{Separation of Variables
Methodology}

\begin{enumerate}[(1)]
    \item Obtain linear PDE for system (with
        boundary conditions and initial
        conditions)
    \item Separate variables to tield decoupled
        ODEs
    \item Impose homogeneous boundary conditions
        to find eigenvalues and eigenfunctions
    \item Use these eigenvalues (constants of
        separation) to find eigenfunctions in the
        other variables.
\end{enumerate}

\subsubsection*{Aside: Solution in characteristic
coordinates}

Recall sine / cosine summation identities which
means our general solution (3.10) becomes
\begin{align*}
    y(x, t)
    &= \half \sum_{n = 1}^\infty \left[ C_n
    \sin \frac{n\pi}{L} (x - ct) + D_n \cos
    \frac{n\pi}{L} (x - ct) \right. \\
    &~~~~~~~~~~~+ \left. C_n \sin \frac{n\pi}{L}
    (x + ct) + D_n \cos \frac{n\pi}{L} (x + ct)
    \right] \\
    &\equiv f(x - ct) + g(x + ct) \tag{3.12}
\end{align*}
The standing wave solution (3.10) is made up of a
right-moving wave (along the characteristic $x -
ct = \eta$, constant) and a left-moving wave ($x +
ct = \xi$, constant) i.e. a general solution with
arbitrary $f$, $g$ (see later). \myskip
Special case: $g(x) = 0$ in (3.1), then $f = g =
\half p$ at $t = 0$.